142 + Interview Questions in PHOTOELECTRIC-EFFECT IN GENERAL AWARENESS Page 2 InterviewSolution

51.	When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. `(hv-phi)` is the maximum kinetic energy of emitted photoelectron (where `phi` is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tuve. Wavelength of charasteristic X-ray depends on the atomic number. Q. If potential difference across the tube is increased thenA. `lamda_(min)` will decreaseB. characteristic wavelength will increaseC. `lamda_(min)` will increaseD. none of these
Answer» Correct Answer - A `lamda_(min)=(hv)/(eV)` `lamda_(min)prop(1)/(V)` As `lamda_(min)` decreases, V increases. Dom choice (a) is correct and the rest are incorrect.

Discussion

52.	The energy reveived from the sun by the earth and surrounding atmosphere is 2 cal`cm^(-2)` "min"`^(-1)` on a surface normal to the rays of sun. Q. What is the total energy received, in joule, by the earth and its atmosphere?A. `10.645xx10^(18)J"min"^(-1)`B. `10.645xx10^(15)J"min"^(-1)`C. `8.645xx10^(17)J"min"^(-1)`D. `9.645xx10^(14)J"min"^(-1)`
Answer» Correct Answer - A The effective area of Earth receiving radiation normally `=pi((D)/(2))^2` `=pi((1.27xx10^(4))/(4))`sq km Energy received by Earth per minute is `(pi)/(4)(1.27)^2xx10^(18)xx2xx4.2` `=10.645xx10^(18)`J `min^(-1)`

Discussion

53.	What is the wavelength of a photon of energy 1 eV?A. `12.4xx10^(3) A`B. `2.4xx10^(3) A`C. `0.4xx10^(2) A`D. `1000A`
Answer» Correct Answer - A `lamda=(hc)/(E)=(6.6xx10^(-34)xx3xx10^(8))/(1xx1.6xx10^(-19))xx10^(10)A` `=12.375xx10^(3)A`

Discussion

54.	An electron and a photon, each has a wavelength of `1.2A`. What is the ratio of their energies?A. 0.048611111111111B. `1:10^(2)`C. `1:10^(3)`D. `1:10^(4)`
Answer» Correct Answer - B `(E_e)/(E_(ph))=((h^2)/(2mlada^2))/((hc)/(lamda))=(h^2)/(2mlamda^2)xx(lamda)/(hc)xx(h)/(2lamdac)` `=(6.6xx10^(-34))/(2xx9.1xx10^(-31)xx1.2xx10^(-10)xx3xx10^(8))` `(1)/(100)`

Discussion

55.	In Q.72, if the velocity of electron is 25% of the velocity of photon, then `(E_e)/(E_(ph))` equalsA. 0.04356B. 0.044C. 0.0472D. 0.125
Answer» Correct Answer - D `(E_e)/(E_(ph))=(4)/(2c)=(1)/(8)`

Discussion

56.	The work function of a substance is `4.0 eV`. The longest wavelength of light that can cause photo electron emission from this substance is approximately. (a) `540 nm` (b ) `400nm` (c ) `310 nm` (d) `220 nm`A. 540 nmB. 400 nmC. 310 nmD. 220 nm
Answer» Correct Answer - C `lamda_(min)=(hc)/(omega)=(6.63xx10^(-34)xx3xx10^(8))/(4(1.6xx10^(-19)))=310xx10^(-9)`

Discussion

57.	A praticle of mass M at rest decays into two particle of masses `m_1` and `m_2`, having non-zero velocities. The ratio of the de Broglie wavelength of the particles `(lamda_1)/(lamda_2)` isA. `(m_1)/(m_2)`B. `(m_2)/(m_1)`C. 1D. `(sqrtm_2)/(sqrtm_1)`
Answer» Correct Answer - C Applying consevation of linear momentum: Initial momentum `=` Final momentum `0=m_1v_1-m_2v_2impliesm_1v_1=m_2v_2` Now, `(lamda_1)/(lamda_2)=((h)/(m_1v_1))/((h)/(m_2v_2))=1`

Discussion

58.	A proton has kinetic energy `E = 100 keV` which is equal to that of a photon . The wavelength of photon is `lambda_(2)`and that of proton is `lambda_(1)` .The ratio of `lambda_(2) // lambda_(1) ` is proportional toA. `E^2`B. `E^((-1)/(2))`C. `E^-1`D. `E^((1)/(2))`
Answer» Correct Answer - B For photon: `E=hv` or `E=(hv)/(lamda)` `implieslamda_2=(hc)/(E)` For proton: `E=(1)/(2)m_pv_p^2` `E=(1)/(2)(m_p^2v_p^2)/(m)impliesP=sqrt(2mE)` `P=(h)/(lamda_1)` `implieslamda_1=(h)/(p)=(h)/(sqrt(2mE))` `(h_2)/(lamda_1)=(hc)/(Exx(h)/(sqrt(2mE)))propE^((-1)/(2))`

Discussion

59.	In a photoelectric experiment anode potential is ploted against plate current. A. A nd B will have different while B and C will have different frequenciesB. B and C will have different intensities while A and C will have different frequenciesC. A and B will have different intensities while A and C will have equal frequenciesD. A and B will have equal intensities while B and C have different frequencies
Answer» Correct Answer - D From the graph, it is clear that A and B have the same stopping potential and therefore the same frequency. Also, B and C have the same intensity.

Discussion

60.	A small mirror of mass m is suspended by a light thread of length l. A short polse of laser falls on the mirror with energy E. Then, whoch of the following statement is correct?A. If the pulse falls normally on the mirror, it deflects by `theta=(2E)/((mcsqrt(2gl)))`B. If the pulse falls normally on the mirror, it deflects by `theta=(2E)/((mcsqrt(2g))`C. Impulse in thread depends on angle at which the pulse falls on the mirrorD. None of the above
Answer» Correct Answer - C Consider energy conservation and concept of impulse.

Discussion

61.	In the intercept on the y-axis is equal toA. `+(W)/(e)`B. `-(W)/(e)`C. `-We`D. `(e)/(W)`
Answer» Correct Answer - B The maximum KE of the photoelectrons is given by `((1)/(2)mv^(2))_(max)=hv-W` Now, `v=c/(lamda)` and ` ((1)/(2)mv^(2))=eV` `eV=(hc)/(lamda)-W` or `V=((hc)/(e))(1)/(lamda)-(W)/(e)` Since V is represented along y-axis and `((1)/(lamda))` along x-axis, the above equation represents a straight line. Slope of straight line `=(hc)/(e)` Intercept of straight line `=-((W)/(e))`

Discussion

62.	If a photocell is illuminated with a radiation of `1240 A`, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength areA. 1 eV,`5200A`B. 2 eV,`6200A`C. 3 eV,`7200A`D. 4 eV,`4200A`
Answer» Correct Answer - B `W=hc-eV_S` `hupsilon=` energy of incident photon Here `hupsilon=(12400)/(1240)eV=10eV` `W=10-8=2eV` So, `lamda_0=` Threshold wavelength `=(12400)/(2eV)A=6200A`

Discussion

63.	A material particle with a rest mass `m_o` is moving with a velocity of light c. Then, the wavelength of the de Broglie wave associated with it isA. `((h)/(m_oc))`B. zeroC. `infty`D. `((m_oc)/(h))`
Answer» Correct Answer - B We know that mass m in motion and the rest mass `m_0` is related through the equation `m=(m_0)/(sqrt(1-(v^2)/(c^2)))` As `v=c`,`m=(m_0)/(sqrt(1-1)=(m_0)/(0)=infty` Therefore, de Broglie wavelength is `lamda=(h)/(mv)=(h)/((infty)(c ))=0`

Discussion

64.	When the energy of the incident radiation is increased by `20 %` , kinetic energy of the photoelectrons emitted from a metal surface increased from `0.5 eV to 0.8 eV`. The work function of the metal isA. `1.0 eV`B. `1.3 eV`C. `1.5 eV`D. `0.65 eV`
Answer» Correct Answer - A `E = W+K` `E =W+0.5` …(i) `1.2 E=W+0.8` …(ii) (ii)/(i) `1.2 =(W+0.8)/(W+0.5)implies 1.2 W+0.6 =W+0.8` `W=1 eV`

Discussion

65.	When ultraviolet radiation is incident on a surface, no photoelectrons are emitted. If another beam causes photoelectrons to be emitted from the surface, it may consist of (i) radio waves (ii) infrared rays (iii) X-rays (iv) gamma raysA. (i),(ii)B. (ii),(iii)C. (i),(iv)D. (iii),(iv)
Answer» Correct Answer - D `upsilon_("gamma")gtupsilon_(X-rays)gtupsilon_("ultraviolet")`

Discussion

66.	the work function of metal is W and `lamda` is the wavelength of the incident radiation. There is no emission of photoelecrons whenA. `lamdagt(hc)/(W)`B. `lamda=(hc)/(W)`C. `lamdalt(hc)/(W)`D. `lamdale(hc)/(W)`
Answer» Correct Answer - A `(hc)/(lamda)ltWimplieslamdagt(hc)/(W)`

Discussion

67.	Which curve shows the relation ship between the energy R and the wavelength `lamda` of a photon of electromagnetic radiation?A. B. C. D.
Answer» Correct Answer - D Each photon has associated with it an energy wave given by `E=hf=(hc)/(lamda)` and graph of E vs. `lamda` is a hyperbola. Thus, `Eprop(1)/(lamda)`

Discussion

68.	Work function of nickel is 5.01 eV. When ultraviolet radiation of wavelength `200A` is incident of it, electrons are emitted. What will be the maximum velocity of emitted electrons?A. `3xx10^(8) ms^(-1)`B. `6.46xx10^(5)ms^(-1)`C. `10.36xx10^(5) ms^(-1)`D. `8.54xx10^(6) ms^(-1)`
Answer» Correct Answer - B Energy corresponding to `2000 A=(12375)/(2000)eV=6.2eV` Maximum kinetic energy is `(6.2-5.01)eV=1.19eV` Now, `(1)/(2)xx9.1xx10^(-31)xxv_(max)^2` `=1.19xx1.6xx10^(-19)` or `v_(max)^2=(1.19xx1.6xx10^(-19)xx2)/(9.1xx10^(-31))` `=.0418xx10^(12)=41.8xx10^(10)` or `v_(max)=6.46xx10^(5)ms^(-1)`

Discussion

69.	Silver has a work function of 4.7 eV. When ultraviolet light of wavelength 100 mm is incident upon it, a potential of 7.7 V is required to stop the photoelectrons from reaching the collector plate. How much potential will be required to stop the photoelectrons when light of wavelength 200mm is incident upon silver?A. 1.5 VB. 3.85 VC. 2.35 VD. 15.4 V
Answer» Correct Answer - A `lamda=(12375)/(E_1(eV))A=1000A` `E_1=12.375eV` Similarly, `(12375)/(lamda_2(A))eV=(12375)/(2000)=6.1875 eV` Now, `E_1W_0=eV_S` and `E_2-W_0=eV_S` Hence, `12.375-W_0=7.7eV` and `6.1875-W_0=eV_s^`` Solving, we get `V_s^`=1.5 V`

Discussion

70.	An image of the sun is formed by a lens, of the focal length of 30 cm, on the metal surface of a photoelectric cell and a photoelectric current I is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photoelectric current in this case isA. `(I)/(2)`B. IC. 2ID. 4I
Answer» Correct Answer - B In both the cases, the intensity is same.

Discussion

71.	If a surface has a work function 4.0 eV, what is the maximum velocity of electrons liberated from the surface when it is irradiated with ultraviolet radiation of wavelength 0.2`mum`?A. `4.4xx10^(5)ms^(-1)`B. `8.8xx10^(7)ms^(-1)`C. `8.8xx10^(5)ms^(-1)`D. `4.4xx10^(7)ms^(-1)`
Answer» Correct Answer - C Let us calculate energy corresponding to `0.2xx10^(-6)`m or `0.2xx10^(-6)xx10^(10)A` or `2000A=(12375)/(2000)=6.1875 eV` `(1)/(2)mv_(max)^(2)=(6.1875-4)eV` or `mv_(max)^(2)=(2xx2.1875xx1.6xx10^(-19))/(9.1xx10^(-31))` or `mv_(max)^(2)=0.769xx10^(12)=0.876xx10^(6)` `=8.76xx10^(5)ms^(-1)`

Discussion

72.	Photoelectric work- function of a metal is `1 eV.` Light of wavelength `lambda = 3000 Å` falls on it. The photoelectrons come out with maximum velocityA. `10ms^(-1)`B. `10^(3)ms^(-1)`C. `10^(4)ms^(-1)`D. `10^(6)ms^(-1)`
Answer» Correct Answer - D `(1)/(2)mv^_(max)^(2)=(hc)/(lamda)-phi_(0)` `(1)/(2)mv_(max)^(2)=(12375eV)/(3000)-1eV` `=3.125xx1.6xx10^(-19)` `v_(max)=sqrt((2xx3.125xx1.6xx10^(-19))/(9.1xx10^(-31)))approx10^(6)ms^(-1)`

Discussion

73.	A photoelectric cell is connected to a source of variable potential difference, connected across it and the photoelectric current resulting `(muA)` is plotted against the applied potential difference (V). The graph in the broken line represents one for a given frequency and intensity of the incident radiation . If the frequency is increased and the intensity is reduced, which of the following graphs of unbroken line represents the new situation?A. AB. BC. CD. D
Answer» Correct Answer - D Intensity reduced therefore saturation current reduced. Frequency increased, therefore stopping potential increased.

Discussion

74.	An electron is accelerated through a potential difference of `200` volts. If `e//m` for the electron be `1.6 xx 10^(11)` coulomb/kg, the velocity acquired by the electron will beA. `8xx10^(6)ms^(-1)`B. `8xx10^(5)ms^(-1)`C. `5.9xx10^(6) ms^(-1)`D. `5.9xx10^(5) ms^(-1)`
Answer» Correct Answer - A `(1)/(2)mv^(2)=eV` or `v+sqrt((2eV)/(m))=sqrt(2xx1.6xx10^(11)xx200)ms^(-1)=8xx10^(6)ms^(-1)`

Discussion

75.	X-rays are used to irradiate sodium and copper surfaces in two separate experiments and stopping potential are determined. The stopping potential isA. equal in both casesB. greater for sodiumC. greater for copperD. infinite in both cases
Answer» Correct Answer - B Sodium has low work funcrion. So, maximum kinetic energy is more in the case of sodium. Thus, stopping potential is more for sodium.

Discussion

76.	When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. `(hv-phi)` is the maximum kinetic energy of emitted photoelectron (where `phi` is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tuve. Wavelength of charasteristic X-ray depends on the atomic number. Q. A monochromatic light is used in a photoelectric experiment on photoelectric effect. The stopping potentialA. is related to the mean wavelengthB. is related to the shortest wavelengthC. is related to the maximum kinetic energy of emitted photoelectronsD. intensity of incident light
Answer» Correct Answer - B::C Stopping potential is the measurement of maximum kinetic energy of emitted protoelectrons and kinetic energy of emitted photoelectrons is linearly related with the frequency of incident light corresponding (i.e., sorresponding to shortest wavelength, KE is maximum). Stopping potential is independent of intensity. From above explanation, it is clear that choices (a) and (d) are

Discussion

77.	The cathode of a photoelectric cell is changed such that the work function changes from `(W_(1) to W_(2) (W_(2) gt W_(1))`. If the current before and after change are `I_(1)` and `I_(2)`, all other conditions remaining unchanged , then (assuming `hv gt W_(2)`)A. `i_(1) = i_(2)`B. `i_(1) lt i_(2)`C. `i_(1) gt i_(2)`D. `i_(1) lt i_(2) lt2i_(1)`
Answer» Correct Answer - A The current depends on intensity

Discussion

78.	A photon of energy E ejects a photoelectron from a metel surface whose work function is `phi_(0)`. If this electron enters into a unifrom magnetic field of induction B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r, is given by, (in the usual notation)A. `sqrt((2m(E - phi))/(eB))`B. `sqrt(2m(E - phi)eB)`C. `sqrt((2e(E - phi))/(mB))`D. `sqrt((2m(E - phi))/(Be))`
Answer» Correct Answer - D `E =phi +K_(max)` `implies K_(max) =E -phi` `R=(mv)/(Bq) =(sqrt(2m K_(max)))/(Be)` `=(sqrt(2m (E -phi)))/(Be)`

Discussion

79.	The electric field associated with a light wave is given by `E= E_0 sin [(1.57 xx 10^7 m^(-1)(x-ct)].` Find the stopping potential when this light is used in an experiment on photoelectric affect with a metal having work - function 1.9 eV.A. `0.6 V`B. `1.2 V`C. `1.8 V`D. `2.4 V`
Answer» Correct Answer - B `E =E_(0) sin (1.57xx10^(7))(x - ct)` `E =E_(0) sin .(omega)/(c)(x - ct)` `(omega)/(c) =1.57xx10^(7)` `(2pi upsilon)/(upsilon lambda) =1.57xx10^(7) =(pi)/(2)xx10^(7)` `lambda =4xx10^(-7) m=400 nm` `(1242)/(lambda) =(1242)/(400) =3.1 eV`

Discussion

80.	The deBroglie wavelength of a particle of kinetic energy K is `lamda`. What would be the wavelength of the particle, if its kinetic energy were `(K)/(4)`?
Answer» If `lamda` is the de Broglie wavelength of the particle of kinetic energy K, then `lamda=(h)/(sqrt(2mK))` Suppose that the de Broglie wavlength of the particle becomes `lamda^``, when its kinetic energy is `(K)/(4)`, then `lamda^`=(h)/(sqrt((2mK)/(4)))=2((h)/(sqrt(2mK)))=2lamda`

Discussion

81.	A helium-neon laser has a power output of 1mW of light of wavelength 632.8 nm. Q. Calculate the energy of each photon in electron volt.
Answer» Here, `lamda=632.8nm=632.8xx10^-9m`, Power, `P=9.42mW=9.42xx10^-3W` a. Now, energy of each photon in the light beam, `E=(hc)/(lamda)=(6.62xx10^-34xx3xx10^8)/(632.8xx10^-9)` `=3.14xx10^-19J=(3.14xx10^-19)/(1.6xx10^-9)=1.963eV` Also, momentum of each photon in the light beam, `p=(h)/(lamda)=(6.62xx10^-34)/(632.8xx10^-9)=1.046xx10^-27kgms^-1` b. The number of photons falling on the target per second, `N=(P)/(E)=(9.42xx10^-3)/(3.14xx10^-19)=3xx10^16s^-1` c. The required speed of the hydrogen atom, `V=(p)/(m_H)=(1.046xx10^-27)/(1.66xx10^-27)=0.63ms^-1`

Discussion

82.	The two lines A and B in fig. shoe the phot of de Broglie wavelength `(lamda)` as a function of `(1)/(sqrtV) `(V is the accelerating potential) for teo particles having the same charge. Which of the two represents the particle of heavier mass?
Answer» In terms of accelerating potential V, the de Broglie wavelength of a charged particle is given by `lamda=(h)/(sqrt(2meV))` …(i) Where e is charge and m is mass of the particle. Equation (i) represents a straight line, whose slope is `(h)/(sqrt(2me))`. The slope of the line is inversely proportional to `sqrtm`. since the slope of line A is lesser, it represents the particle of heavier mass

Discussion

83.	Photoelectric effect experiments are performed using three different metal plates `p,q` and`r` having work function `phi_(p) = 2.0 eV, phi_(e) = 2.5 eV and phi_(r) = 3.0 eV` respectively A light beam containing wavelength of `550nm , 450 nm` and `350nm ` with equal intensities illuminates each of the plates . The correct `I -V` graph for the experiment is [Take hc = 1240 eV nm]A. B. C. D.
Answer» Correct Answer - A `E_(lambda1) =550 nm =(1242)/(550)eV =2.25 eV` `E_(lambda2) =450 nm =(1240)/(450)eV =2.8 eV` `E_(lambda3) =350 nm =(1242)/(350)eV =3.5 eV` From plate p, electrons are ejected by wave lengths 550, 450 and 350 nm From plate q electrons are ejected by wavelength 450 and 350 nm From plate r electrons are ejected by wave length 350 nm

Discussion

84.	The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are increased by a factor of two. This willA. increase the maximum kinetic energy, the photoelectrons, as well as photoelectric current by a factor of 2B. increase the maximum kinetic energy of the photoelectrons and would increase the photoelectric current by a factor of 2C. increase the maximum kinetic energy of the photoelectrons by a factor of 2 and will gave no effect on the magnitude of the photoelectric current producedD. not produce any effect on the kinetic energy of the emitted electrins but will increase the photoelectric current by a factor of 2
Answer» Correct Answer - B `((1)/(2)mv^(2))_(max)=hv-W` When `upsilon` is doubled (W remains same), `((1)/(2)mv^(2))_(max)` i.e., (KE) is increased. The photoelectric current is directly proportional to the intensity of incident light.

Discussion

85.	The eye can detect `5xx10^(4)` photons `(m^2s)^(-1)` of green light (`lamda=5000A)`, whole ear can detect `10^(-13)Wm^(2)` . As a power detector, which is more sensitive and by what factor?A. Eye is more sensitive and by a factor of 5.00B. Ear is more sensitive by a factor of 5.00C. Both are equally sensitiveD. Eye is more sensitive by a factor of `10^(-1)`
Answer» Correct Answer - A Energy received by the eye, `E=(nhc)/(lamda)` `=(5xx10^(94)xx6.610^(-34)xx3xx10^(8))/(5000xx10^(-10))` `=0.2xx10^(-13)Wm^(-2)` So, eye in more sensitive by a factor of `(1)/(0.200)=5.00`

Discussion

86.	`10^-3` W of `5000A` light is directed on a photoelectric cell. If the current in the cell is `0.16muA`, the percentage of incident photons which produce photoelectrons, isA. `40%`B. `0.04%`C. `20%`D. `10%`
Answer» Correct Answer - B Number of photons falling per second : `N_p=(10^-3)/((6.6xx10^-34xx3xx10^8)/(5000xx10^-10))=2.5xx10^15` Let `N_e` is the number of photoelectrons emitted per second. `I=(q)/(t)=(N_ee)/(1)impliesN-e=(I)/(e)=(0.16xx10^-6)/(1.6xx10^-19)=10^12` percentage of photons producing photoelectrons `=(N_e)/(N_p)xx100=(10^12)/(2.5xx10^15)xx100=0.04%`

Discussion

87.	When light of sufficiently high frequency is incident on a metallic surface, electrons are emitted from the metallic surface. This phenomenon is called photoelectric emission. Kinetic energy of the emitted photoelectrons depends on the wavelength of incident light and is independent of the intensity of light. Number of emitted photoelectrons depends on intensity. `(hv-phi)` is the maximum kinetic energy of emitted photoelectron (where `phi` is the work function of metallic surface). Reverse effect of photo emission produces X-ray. X-ray is not deflected by electric and magnetic fields. Wavelength of a continuous X-ray depends on potential difference across the tuve. Wavelength of charasteristic X-ray depends on the atomic number. Q. If frequency `(upsilongtupsilon_0)` of incident light becomes n times the initial frequency (v), then KE of the emitted photoelectrons becomes (`v_0` threshold frequency).A. n times of the initial kinetic energyB. more than n times of the intial kinetic energyC. less than n times of the initial kinetic energyD. kinetic energy of the emitted photoelectrons remains unchanged
Answer» Correct Answer - B `KW_1=hv-phi` `KE_2=nhv-phi` `=n(hv-phi)+(n-1)phi` `KE_2=nKE_1+(n-1)phi` `KE_2ltnKE_1`

Discussion

88.	The work function of Silver and sodium are `4.6 and 2.3 eV, ` respectively . The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is
Answer» Slope of graph is `(h)/(e)=`constant `implies`1

Discussion

89.	Photoelectric effect supports quantum nature of light because (a) there is a minimum frequency of light below which no photo electrons are emitted (b) the maximum kinetic energy of photo electrons depends only on the frequency of light and not on its intensity (c ) even when the metal surface is faintly illuminated, the photo electrons leave the surface immediately (d) electric charge of the photo electrons is quantisedA. there is a minimum frequency of light below which no photoelectrons are emittedB. the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensityC. even when the metal surface is faintly illuminated, the photoelectrons leave the surface immediatelyD. electric charge of the photoelectrons is quantized
Answer» Correct Answer - A::B::C Standard result

Discussion

90.	If the kinetic energy of the particle is increased to `16` times its previous value , the percentage change in the de - Broglie wavelength of the particle isA. 75B. 60C. 50D. 25
Answer» Correct Answer - A de-Broglie wavelength `lambda =(h)/(p) =(h)/(sqrt(2mK))` `(lambda_(2))/(lambda_(1)) =sqrt((K_(1))/(K_(2))) =sqrt((K)/(16K)) =(1)/(4)` `%` changed `=((lambda_(2))/(lambda_(1))-1)xx100= ((1)/(4)-1)xx100=-75%` i.e. `75%` decrease

Discussion

91.	The figure shows a plot of photo current versus anode potential for a photosensitive surface for three different radiations. Which one of the following is a correct statement ? A. curves a and b represent incident radiations of different frequencies and different intensitiesB. curves a and b represents incident radiations of same frequency but of different intensitiesC. curves b and c represents incident radiations of different frequencies and different intensitiesD. curves b and c represents incident radiations of same frequency having same intensity
Answer» Correct Answer - B Stopping potential is same for a and b, hence their frequencies are same. The maximum current values are different for a and b so they will have different intensities

Discussion

92.	An element of atomic number 9 emits `K_(alpha)` X-ray of wavelength `lamda`. Find the atomic number of the element which emits `K_(alpha)` X-ray of wavelength `4lamda`.
Answer» For `K_(alpha)` X-ray, `(Z-1)^2lamda=`constant. Hence, `(9-1)^2lamda=(Z-1)^2(4lamda)` `(Z-1)^2=(64)/(4)=16` `Z-1=4` or `Z=5`

Discussion

93.	In the arrangement shown if Fig. `y=1.0mm`,`d=0.24mm` and `D=1.2m`. The work function of the material of the emitter if 2.2 eV. Find the stopping potential V needed to stop photocurrent (in`xx10^(-1)`V).
Answer» Given: Fringe width, `y=1.0mmxx2=2.0mm` `d=0.24mm`,`W_0=2.2eV`,`D=1.2m` `y=(lamdaD)/(d)` or `lamda=(yd)/(D)` `=(2xx10^(-3)xx0.24xx10^(-3))/(1.2)=4xx10^(-7)`m `E=(hc)/(lamda)=(4.14xx10^(-15)xx3xx10^(8))/(4xx10^(-7))=3.105 eV` Stopping potential `eV_0=3.105-2.2=0.905 eV` `V_0=(0.905)/(1.6xx10^(-19))xx1.6xx10^(-19)V=0.905V`

Discussion

94.	An experimental setup of verification of photoelectric effect is shown if Fig. The voltage across the electrodes is measured with the help of an ideal voltmeter, and which can be varied by moving jockey J on the potentiometer with wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is `2 Omega`. The resistance of 100 cm long potentiometer wire is `8 Omega`. The photocurrent is measured with the help of an ideal ammeter. Two plates of potassium oxide of area `50 cm^2` at separation 0.5mm are used in the vacuum tube. Photocurrent in the circuit is very small, so we can treat the potentiometer circuit as an independent circuit. The wavelength of various colors is as follows: Q. When other light falls on the anode plate, the ammeter reading remains zero till jockey is moved from the end P to the middle point of the wire PQ. Thereafter, the deflection is recorded in the ammeter. The maximum kinetic energy of the emitted electron isA. 16 eVB. 8 eVC. 4 eVD. 10 eV
Answer» Correct Answer - B Stopping potential, `V_(S)=8V` and `KE=eV_(S)` `KE=8eV`

Discussion

95.	The maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the ………. Of the incident radiation .
Answer» According to laws of photoelectric effect, `(1)/(2)mv^2=hv-hv_0` i.e., the maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the frequency of incident radiation.

Discussion

96.	Electrons taveling at a velocity of `2.4xx10^(6) ms^(-1)` enter a region of crossed electric and magnetic fields as shoen in Fig. If the electric field is `3.0xx10^(6)` Vm and the flux density of the magnetic field is 1.5 T, the electrons upon entering the region of the crossed fields willA. continue to travel undeflected in their original derection.B. be deflected upward in the plane of the diagramC. be deflected upward on the plane of the diagramD. none of the above
Answer» Correct Answer - C The ratio of electric force to magnetic force is: `(F_e)/(F_B)=(qE)/(qvB)=(3xx10^(6))/(2.4xx10^(6)xx1.5)=(5)/(6)` Now, magnetic force on beta particles acts downward, whereas electric force acts upward and both are in the plane of diagram. But since magnetic force is larger, so beta particles are deflected downward.

Discussion

97.	When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic are N and T respectively. If the intensity of radiation is 2 I, the number of emitted electrons and their maximum kinetic energy are respectively.A. N and 2TB. 2N and TC. 2N and 2TD. N and T
Answer» Correct Answer - B Number of electrons `prop` intensity `K_(max)` does not depends on intensity

Discussion

98.	The threshold frequency for certain metal is `v_0`. When light of frequency `2v_0` is incident on it, the maximum velocity of photoelectrons is `4xx10^(6) ms^(-1)`. If the frequency of incident radiation is increaed to `5v_0`, then the maximum velocity of photoelectrons will beA. `4xx10^(6) m//sec`B. `6xx10^(6) m//sec`C. `8xx10^(6) m//sec`D. `16xx10^(6) m//sec`
Answer» Correct Answer - C `h.2 upsilon_(0) = h upsilon_(0) +(1)/(2)m(4xx10^(6))^(2)` ….(i) `h.5 upsilon_(0) = h upsilon_(0) +(1)/(2)m v_(max)^(2)` …..(ii) `(v_(max)^(2))/((4xx10^(6))^(2)) =4 implies v_(max) =8xx10^(6) m//sec`

Discussion

99.	An experimental setup of verification of photoelectric effect is shown if Fig. The voltage across the electrodes is measured with the help of an ideal voltmeter, and which can be varied by moving jockey J on the potentiometer with wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is `2 Omega`. The resistance of 100 cm long potentiometer wire is `8 Omega`. The photocurrent is measured with the help of an ideal ammeter. Two plates of potassium oxide of area `50 cm^2` at separation 0.5mm are used in the vacuum tube. Photocurrent in the circuit is very small, so we can treat the potentiometer circuit as an independent circuit. The wavelength of various colors is as follows: Q. Calculate the number of electrons that appear on the surface of the cathode plate, when the jockey is connected at the end P of the potentiometer wire. Assume that no radiation is falling on the plates.A. `8.85xx10^(6)`B. `11.0625xx10^(9)`C. `8.85xx10^(9)`D. `0`
Answer» Correct Answer - C `Q=Cvimplies` `"ne"=(epsi_0A)/(d)V` `n=(2.85xx10^(-12)xx10)/(0.5xx10^(-3)xx1.6xx10^(-19))xx16` `n=8.85xx10^(9)`

Discussion

100.	An experimental setup of verification of photoelectric effect is shown if Fig. The voltage across the electrodes is measured with the help of an ideal voltmeter, and which can be varied by moving jockey J on the potentiometer with wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is `2 Omega`. The resistance of 100 cm long potentiometer wire is `8 Omega`. The photocurrent is measured with the help of an ideal ammeter. Two plates of potassium oxide of area `50 cm^2` at separation 0.5mm are used in the vacuum tube. Photocurrent in the circuit is very small, so we can treat the potentiometer circuit as an independent circuit. The wavelength of various colors is as follows: Q. Which of the following colors may not give photoelectric effect for this cathode?A. GreenB. VioletC. RedD. Orange
Answer» Correct Answer - C The range of wavelength for red light is beyond the wavelength of incident light.

Discussion

Explore topic-wise InterviewSolutions in .

What is the wavelength of a photon of energy 1 eV?A. `12.4xx10^(3) A`B. `2.4xx10^(3) A`C. `0.4xx10^(2) A`D. `1000A`

An electron and a photon, each has a wavelength of `1.2A`. What is the ratio of their energies?A. 0.048611111111111B. `1:10^(2)`C. `1:10^(3)`D. `1:10^(4)`

In Q.72, if the velocity of electron is 25% of the velocity of photon, then `(E_e)/(E_(ph))` equalsA. 0.04356B. 0.044C. 0.0472D. 0.125

The work function of a substance is `4.0 eV`. The longest wavelength of light that can cause photo electron emission from this substance is approximately. (a) `540 nm` (b ) `400nm` (c ) `310 nm` (d) `220 nm`A. 540 nmB. 400 nmC. 310 nmD. 220 nm

A praticle of mass M at rest decays into two particle of masses `m_1` and `m_2`, having non-zero velocities. The ratio of the de Broglie wavelength of the particles `(lamda_1)/(lamda_2)` isA. `(m_1)/(m_2)`B. `(m_2)/(m_1)`C. 1D. `(sqrtm_2)/(sqrtm_1)`

A proton has kinetic energy `E = 100 keV` which is equal to that of a photon . The wavelength of photon is `lambda_(2)`and that of proton is `lambda_(1)` .The ratio of `lambda_(2) // lambda_(1) ` is proportional toA. `E^2`B. `E^((-1)/(2))`C. `E^-1`D. `E^((1)/(2))`

In the intercept on the y-axis is equal toA. `+(W)/(e)`B. `-(W)/(e)`C. `-We`D. `(e)/(W)`

If a photocell is illuminated with a radiation of `1240 A`, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength areA. 1 eV,`5200A`B. 2 eV,`6200A`C. 3 eV,`7200A`D. 4 eV,`4200A`

A material particle with a rest mass `m_o` is moving with a velocity of light c. Then, the wavelength of the de Broglie wave associated with it isA. `((h)/(m_oc))`B. zeroC. `infty`D. `((m_oc)/(h))`

When the energy of the incident radiation is increased by `20 %` , kinetic energy of the photoelectrons emitted from a metal surface increased from `0.5 eV to 0.8 eV`. The work function of the metal isA. `1.0 eV`B. `1.3 eV`C. `1.5 eV`D. `0.65 eV`

When ultraviolet radiation is incident on a surface, no photoelectrons are emitted. If another beam causes photoelectrons to be emitted from the surface, it may consist of (i) radio waves (ii) infrared rays (iii) X-rays (iv) gamma raysA. (i),(ii)B. (ii),(iii)C. (i),(iv)D. (iii),(iv)

the work function of metal is W and `lamda` is the wavelength of the incident radiation. There is no emission of photoelecrons whenA. `lamdagt(hc)/(W)`B. `lamda=(hc)/(W)`C. `lamdalt(hc)/(W)`D. `lamdale(hc)/(W)`

Which curve shows the relation ship between the energy R and the wavelength `lamda` of a photon of electromagnetic radiation?A. B. C. D.

Work function of nickel is 5.01 eV. When ultraviolet radiation of wavelength `200A` is incident of it, electrons are emitted. What will be the maximum velocity of emitted electrons?A. `3xx10^(8) ms^(-1)`B. `6.46xx10^(5)ms^(-1)`C. `10.36xx10^(5) ms^(-1)`D. `8.54xx10^(6) ms^(-1)`

If a surface has a work function 4.0 eV, what is the maximum velocity of electrons liberated from the surface when it is irradiated with ultraviolet radiation of wavelength 0.2`mum`?A. `4.4xx10^(5)ms^(-1)`B. `8.8xx10^(7)ms^(-1)`C. `8.8xx10^(5)ms^(-1)`D. `4.4xx10^(7)ms^(-1)`

Photoelectric work- function of a metal is `1 eV.` Light of wavelength `lambda = 3000 Å` falls on it. The photoelectrons come out with maximum velocityA. `10ms^(-1)`B. `10^(3)ms^(-1)`C. `10^(4)ms^(-1)`D. `10^(6)ms^(-1)`

An electron is accelerated through a potential difference of `200` volts. If `e//m` for the electron be `1.6 xx 10^(11)` coulomb/kg, the velocity acquired by the electron will beA. `8xx10^(6)ms^(-1)`B. `8xx10^(5)ms^(-1)`C. `5.9xx10^(6) ms^(-1)`D. `5.9xx10^(5) ms^(-1)`

X-rays are used to irradiate sodium and copper surfaces in two separate experiments and stopping potential are determined. The stopping potential isA. equal in both casesB. greater for sodiumC. greater for copperD. infinite in both cases

The electric field associated with a light wave is given by `E= E_0 sin [(1.57 xx 10^7 m^(-1)(x-ct)].` Find the stopping potential when this light is used in an experiment on photoelectric affect with a metal having work - function 1.9 eV.A. `0.6 V`B. `1.2 V`C. `1.8 V`D. `2.4 V`

The deBroglie wavelength of a particle of kinetic energy K is `lamda`. What would be the wavelength of the particle, if its kinetic energy were `(K)/(4)`?

A helium-neon laser has a power output of 1mW of light of wavelength 632.8 nm. Q. Calculate the energy of each photon in electron volt.

The two lines A and B in fig. shoe the phot of de Broglie wavelength `(lamda)` as a function of `(1)/(sqrtV) `(V is the accelerating potential) for teo particles having the same charge. Which of the two represents the particle of heavier mass?

`10^-3` W of `5000A` light is directed on a photoelectric cell. If the current in the cell is `0.16muA`, the percentage of incident photons which produce photoelectrons, isA. `40%`B. `0.04%`C. `20%`D. `10%`

The work function of Silver and sodium are `4.6 and 2.3 eV, ` respectively . The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is

If the kinetic energy of the particle is increased to `16` times its previous value , the percentage change in the de - Broglie wavelength of the particle isA. 75B. 60C. 50D. 25

An element of atomic number 9 emits `K_(alpha)` X-ray of wavelength `lamda`. Find the atomic number of the element which emits `K_(alpha)` X-ray of wavelength `4lamda`.

In the arrangement shown if Fig. `y=1.0mm`,`d=0.24mm` and `D=1.2m`. The work function of the material of the emitter if 2.2 eV. Find the stopping potential V needed to stop photocurrent (in`xx10^(-1)`V).

The maximum kinetic energy of electrons emitted in the photoelectric effect is linearly dependent on the ………. Of the incident radiation .