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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
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. It is found that ammeter current remains unchanged `(2muA)` even when the jockey is moved from the end P to the middle point of the potentiometer wire. Assuming that all the incident photons eject electrons and the power of the light incident is `4xx10^(-6)` W. Then, the color of the incident light isA. GreenB. VioletC. RedD. Orange |
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Answer» Correct Answer - D `P=(nfC)/(elamda)` Where `n=` no. of photons incident per unit time. Also, `I="ne"` `impliesP=(IhC)/(elamda)` `lamda=((2xx10^(-6))(6.6xx10^(-34))(3xx10^(8)))/((4xx10^(-6))(1.6xx10^(-19)))` `=(9.9)/(1.6)xx10^(-7)m=(9900)/(1.6)A` `=6187A` Which is in the range of orange light. |
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| 102. |
Photoelectric effect supports quantum nature of light becauseA. there is minimum frequency of light below which no photoelectrons are emittedB. the maximum KE of photoelectrons depends only on the frequency of light and not on its intensityC. even when the metal surface is faintly illuminated by light of wavelength less than the threshold wavelength, the photoelectrons leave the surface immediatelyD. electric charge of photoelectrons is quantized |
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Answer» Correct Answer - A::B::C Existence of cut off frequency and photoemission takes place even when intensity is low. |
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| 103. |
In a photoelectric emission process the maximum energy of the photo - electrons increase with increasing intensity of the incident light . |
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Answer» False. `(KE)_(max)=hv-hv_0implies(KE)_(max)propv` Thus, maximum kinetic energy is proportional to frequency and not intensity. |
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| 104. |
A metal surface in an evacuated tube is illuminated with monochromatic light causeng the emission of photoelectrons which are collected at an adjacent electrode. For a given intensity of light, the in which the photocurrent I depends in the potential difference V between the electrodes is shown by approximate graphi in Fig. If the experiment were repeated with light of twice the intensit but the same wavelength, which of the graphs below would best represent the new relation between I and V? (In these graphs, the result of the original experiment is idicated by a broken line.)A. B. C. D. |
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Answer» Correct Answer - B On increaseng intensity, only saturation current increases, whereas retarding potential remains the same because wavelength of lighti s unchanged. |
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| 105. |
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. `(4)/(5)xx10^(6)ms^(-1)`B. `2xx10^(6)ms^(-1)`C. `8xx10^(6)ms^(-1)`D. `2xx10^(7)ms^(-1)` |
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Answer» Correct Answer - C In the first case, `(1)/(2)mv_(max)^(2)=2hupsilon_0-hupsilon_0=hupsilon_0` In the second case, `(1)/(2)mv_(max)^(2)=5hupsilon_0-hupsilon_0=4hupsilon_0` Clearly, `v_(max)` is doubled. |
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| 106. |
A point source of light is taken away from the experimental setup of photoelectric effect. For this situation, mark out the correct statement (s).A. Saturation photocurrent decreasesB. Saturation photocurrent increasesC. Stopping potential remains the sameD. Stopping potential increases |
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Answer» Correct Answer - A::C As the source is taken away, the intensitiy of light reaching the target decreases, and hence the photocurrent decreases. But as motion of the source does not affect frequency of light the stopping potential given by `V_0=((hc)/(e))-((phi)/(e))` remains the same |
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| 107. |
In a photoelectric emission, electrons are ejected from metals X and Y by light of frequency f. The potential difference V required to stop the electrons is measured for various frequencies.. If Y has a greater work function than X, which graph illustrates the expected results?A. B. C. D. |
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Answer» Correct Answer - A V versus F has a constant slope of `(h)/(e)`, so both lines must be parallel. Also, work function is equal to intercept on f-axis |
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| 108. |
If stopping potentials corresponding to wavelengths `4000A` and `4500A` are 1.3 V and 0.9 V, respectively, then the work function of the metal isA. 0.3 eVB. 1.3 eVC. 2.3 eVD. 5 eV |
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Answer» Correct Answer - C `eV_S=(hc)/(lamda)-phi` or `eV_S+phi_0=(hc)/(lamda)` or `lamda=(hc)/(eV_S+phi_0)` `implies(lamda_2)/(lamda_1)=(eV_(S1)+phi_0)/(eV_(S2)+phi_0)` or `(4500)/(4000)=(1.3+phi_0)/(0.9+phi_0) ` or `phi_0=851.-9.xx0.9` Solving: `phi_0=(10.4-8.1)=2.3eV` |
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| 109. |
In a photoelectric experiment, the wavelength of the incident light is decreased from `6000A` to `4000A`. While the intensity of radiations remains the same,A. the cut off potential will decreaseB. the cut off potential will increaseC. the photoelectric current will increaseD. the kinetic energy of the emitted electrons will increase |
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Answer» Correct Answer - B::D `eV_0=E_K^(max)=(hc)/(lamda)-W` |
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| 110. |
A 100 W point source emits monochromatic light of wavelength `6000 A` Q. Calculate the photon flux (in SI unit) at a distance of 5m from the source. Given `h=6.6xx10^(34)`J s and `c=3xx10^(8)ms^(-1)`A. `10^(15)`B. `10^(18)`C. `10^(20)`D. `10^(22)` |
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Answer» Correct Answer - B Intensity (i.e., energy flux) at a distance of 5m from the source of power P, `I=(P)/(4pir^2)=(100)/(4pixx5^2)=(1)/(pi)(W)/(m^2)` The number of photons passing normally per unit area per second i.e., Photon flux `=(I)/(E)=((1)/(pi))/(3.3xx10^(-19))approx10^(18)` photons `m^(-2)s^(-1)` |
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| 111. |
In a photoelectric cell, the wavelength of incident light is chaged from `4000A` to `3600A`. The change in stopping potential will beA. 0.14 VB. 0.24 VC. 0.35 VD. 0.44 V |
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Answer» Correct Answer - C `eV_S=hupsilon-phi_0` `eV_s^`=hupsilon^`-phi_0` `e(V_S^`-V_S)=hupsilon^`-hupsilon=((12374)/(3600)-(12375)/(4000))` `V_S^`-V_S=3.44-3.09=0.35V` |
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| 112. |
A 100 W point source emits monochromatic light of wavelength `6000 A` Q. Calculate the total number of photons emitted by the source per second.A. `5xx10^(20)`B. `8xx10^(20)`C. `6xx10^(21)`D. `3xx10^(20)` |
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Answer» Correct Answer - D Energy of a photon of wavelength `lamda`, `E=hv=((hc)/(lamda))` `E=((6.6xx10^(-34))xx(3xx10^(8)))/(6000xx10^(-10))` `=3.3xx10^(-19)J` So, if n is the number of photons emitted per second, `nE=("energy")/("second")="power"(P)` Hence, `n=(P)/(E)=(100)/(3.3xx10^(-19))` `approx3xx10^(20)`, photons `s^(-1)` |
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| 113. |
A 100 W point source emits monochromatic light of wavelength `6000 A` Q. 1.5 mW of `4000A` light is directed at a photoelectric cell If 0.10 per cent of the incident photons produce photoelectrons, find current is the cell. [Given `h=6.6xx10^(-34)Js`,`c=3xx10^(8)ms^(-1)` and `e=1.6xx10^(-19)C`]A. `0.59muA`B. `1.16 muA`C. `0.48muA`D. `0.79muA` |
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Answer» Correct Answer - C Energy of photon `E=hv(hc)/(lamda)=(6.6xx10^(-34)xx3xx10^(8))/(4000xx10^(-10))` `approx5xx10^(-19)J` So, number of photons emitted per second by light source, `n_p=(P)/(E)=(1.5xx10^-3)/(5xx10^(-19))=3xx10^(15)` Now, as only 0.1% of the photons emit electrons, number of photoelectrons emitted per second, `n_e=(0.1)/(100)xx3xx10^(15)=3xx10^(12)` But current is the rate of flow of charge i.e., `(I)=(q)/(t)=(Ne)/(t)-n_exxe[as(N)/(t)=n_e]` So, `I=3xx10^(12)xx1.6xx10^(-19)=0.48muA` |
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| 114. |
Determine the de Broglie wavelength of a proton, whose kinetice energy is equal to the rest of the mass energy of an electron. Given that the mass of an electron is `9xx10^(-31)` kg and the mass of a proton is `1837` times as that of the electron. |
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Answer» Here mass of electron, `m_0=9.1xx10^(-31)`kg Mass of proton, `m=1837xx9.1xx10^(-31)=1.67xx10^(-27)`kg let v be the velocity of the proton. Then, `(1)/(2)mv^(2)=m_0c^(2)` or `m^(2)v^(2)=2mm_0c^(2)` or `mv=sqrt(2mm_0c)` `=sqrt(2xx1.67xx10^(-27)xx9.1xx10^(31))xx3xx10^(8)` `=1.654xx10^(20)kgms^(-1)` Threfore, de Broglie wavelength of the proton, `lamda=(h)/(mv)=(6.62xx10^(-32))/(1.654xx10^(-20))=4xx10^(-14)m` |
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| 115. |
An electron and a photon possess the same de Broglie wavelength. If `E_e` and `E_ph` are, respectively, the energies of electron and photon while v and c are their respective velocities, then `(E_e)/(E_(ph))` is equal toA. `(v)/(c )`B. `(v)/(2c)`C. `(v)/(3c)`D. `(v)/(4c)` |
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Answer» Correct Answer - B `lamda=(h)/(sqrt(2mE_e))=(hc)/(E_(ph))` or `2mE_e=(E_(ph)^(2))/(c^(2))` but `E_(e)=(1)/(2)mv^(2)` or `m=(2E_(e))/(v^(2))` `2[(2E)/(v^(2))]E_(e)=(E_(ph)^(2))/(c^(2))` or `(4E_e^2)/(v^2)=(E_(ph)^2)/(c^2)` or `(E_e^2)/(E_(ph)^(2))=(v^2)/(4c^2)` or `(E_(e))/(E_(ph))=(v)/(2c)` |
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| 116. |
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 radiated, in J `min^(-1)`, by the sun to the universe? Distance of the sun from the earth is `1.49xx10^(11)`m.A. `2.3444xx10^(28)J"min"^(-1)`B. `2.33xx10^(24)J"min"^(-1)`C. `2.34xx10^(20)J"min"^(-1)`D. none of these |
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Answer» Correct Answer - A The area of surface surrounding sun at a distance equal to earth distance `4pid^2` Energy raditad by sun (in J `"min"^(-1))` is `2xx4.2[4pi(1.49)^2xx10^(26)]`J`"min"^(-1)` `=2.3444xx10^(28)` J `"min"^(-1)` |
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| 117. |
What is the energy of a proton possessing wavelength `0.4A`?A. 0.51 eVB. 1.51 eVC. 10.51 eVD. 100.51 eV |
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Answer» Correct Answer - A `lamda=(0.286)/(sqrtE(i n eV))A` `sqrtE(i n eV)=(0.286)/(0.4)approx0.707approx(1)/(sqrt2)` E(in eV) `=0.51` |
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| 118. |
A toy truck has dimensions as shown in fig. and its width, normal to the plane of this paper, is d. The sun rays are incident on it as shiwn in the figure. If intensity of rays is I and all surfaces of truck are perfectly black, calculate tension in the thread used to keep the truck stationary. Neglect friction. |
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Answer» First, we have to calculate power incident on the truck and then momentum of light photons incident per second. Since the truck surfaces are perfectly black, therefore momentum finally reduces to zero. It means that the rate of change of momentum of light is equal to the momentum of photons incident per second. Area of top of the driving cabin`=bd`. Light is incident due to the component I `costheta` of intensity I. Power incident on top of the cabin`=bdIcostheta` Similarly, power incident on rear horizontal part of the truck is `adIcostheta` and power incident on rear vertical wall of the driving cabin in `hdIsintheta`. Total power incident on the truck is `P=(bcostheta+acostheta+hsintheta)Id` Momentum of these photons is `p=(P)/(c)` (where c is speed of light in vacuum). But rate of change in momentum of photons, `p=((Id)/(c))(bcostheta+acostheta+hsintheta)` This momentum is inclined at angle `theta` with vertical. Its vertical component `pcostheta` is balanced by normal reaction of the floor and horizontal component `psintheta` is balanced by tension in the thread. Tension `=p sintheta=((Id)/(c)(bcostheta+acostheta+hsintheta)sintheta)` |
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| 119. |
What voltage must be applied to an elect-non microscope to produce electrons of wavelength `0.4A`? |
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Answer» Here `lamda=0.4A=0.4xx10^(-10)m` Let V be the required voltage. Then, `lamda=(h)/(sqrt(2meV))` or `lamda=(12.27)/(sqrtV)` (in A) `0.4=(12.27)/(sqrtV)` or `sqrtV=30.675` or `V=(30.675)^(2)=940.96V` |
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| 120. |
An electron is accelerated by a potential difference of 50 V. Find the de Broglie wavelength associated with it. |
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Answer» For an electron, de Broglie wavelength is given by `lamda=sqrt((150)/(V))=sqrt((150)/(50))=1.73A` |
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| 121. |
An electron is accelerated by a potential difference of 25 V. Find the de Broglie wavelength associated with it. |
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Answer» For and electron, de Broglie wavelength is given by `lamda=sqrt((150)/(V))=sqrt((150)/(25))=sqrt6A` `approx2.5A` |
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| 122. |
Find the de Broglie wavelength of 2 MeV proton. Mass of proton `=1.64xx10^(-27)kg`,`h=6.625xx10^(-34)Js` |
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Answer» Energy of proton, `E=2MeV=2xx1.6xx10^-13J` de Broglie wavelength, `lamda=(h)/(p)=(h)/(sqrt(2mE))` `implieslamda=(6.625xx10^-34)/(sqrt(2xx1.64xx10^-27xx1.6xx10^-13))` `implieslamda=2.044xx10^-14m` |
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| 123. |
Calculate the valocity of a photoelectron if the work function of the target material is `1.24eV` and the wavelength of incident light is `4.36xx10^(-7)`m. What retarding potentia is necessary to stop the emission of the electrons? |
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Answer» We know that `(1)/(2)mv^(2)=hf-phi` or `(1)/(2)mv^(2)=(hf)/(lamda)-phi` Subsituting the values, we get `v=7.43xx10^(5)ms^(-1)`. Let `V_S` be the retarding potential required to stop the emissions of photoelectrons. Then, we have `eV_s=(1)/(2)mv^(2)` `impliesV_S=((1)/(2)mv^(2))/(e)=(2.511xx10^(-19))/(1.6xx10^(-19))=1.58V` |
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| 124. |
Which of the following graphs correctly represents the variation of particle momentum with associated de Broglie wavelength?A. B. C. D. |
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Answer» Correct Answer - D `lamda=(h)/(p)implieslamdaprop(1)/(p)` So, graph between `lamda` and p is a rectangular hyperbola. |
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| 125. |
The work function for tungsten and sodium are `4.5 eV` and `2.3 eV` respectively . If the threshold wavelength `lambda` for sodium is `5460 Å`, the value of `lambda` for tungsten isA. `5893 A`B. `10683 A`C. `2791 A`D. `528 A` |
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Answer» Correct Answer - C `phi_0=hupsilon_0=(hc)/(lamda_0)` `phi_0lamda_0=constant` `4.5xxlamda=2.3xx5460` or `lamda=(2.3xx5360)/(4.5)A=2790A approx2791A` |
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| 126. |
The work function of a metallic surface is 5.01 eV. The photoelectrons are emitted when light of wavelength `2000A` falls on it. The potential difference applied to stop the fastes photoelectrons is `[h=4.14xx10^(-15)eVs]`A. 1.2 VB. 2.24 VC. 3.6 VD. 4.8 V |
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Answer» Correct Answer - A `eV=hv-phi_0` `=((12375)/(2000)-5.01)eV` `V=(6.1875-5.01)V-1.18Vapprox1.2V` |
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| 127. |
When a surface is irradiated with light of wavelength `4950A`, a photocurrent appears which vanishes if a retarding potential greater than `0.6V` is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to `1.1V`. Find the work function of the emitting surface and the wavelength of the second source. |
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Answer» We know that `hf_1=phi+(1)/(2)mv_1^(2)` and `(1)/(2)mv_1^2=eV_1` `phi=hf_1-eV_1=(hc)/(lamda_1)-eV=((6.6xx10^(-34))xx(3xx10^(8)))/(4950xx10^(-10))-(1.6xx10^(-19))(0.6)` `=3.04xx10^(-19)V`,`hf_2=phi+eV_2` `(hc)/(lamda_2)=3.04xx10^(-19)+(1.6xx10^(-19))xx1.1=4.8xx10^(-19)` `implieslamda_2=((6.6xx10^(34))xx(3xx10^(8)))/(4.8xx10^(-19))=4125A` |
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| 128. |
Five volt of stopping potential is needed for the photoelectrons emitted out of a surface of work function 2.2 eV by the radiation of wavelengthA. `1719A`B. `3444A`C. `861A`D. `3000A` |
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Answer» Correct Answer - A `hv=5eV+2.2eV=7.2eV` `7.2=(12375)/(lamda("in" A))` or `lamda("in" A)=(12375)/(7.2)approx1719` |
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| 129. |
The stopping potential for photoelectrons emitted from a surface illuminated by light wavelength of `5893A` is `036 V`. Calculate the maximum kinetic energy of photoelectrons, the work function of the surface, and the threshold frequency. |
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Answer» We know that `KE_(max)=hf-phi((hc)/(lamda))-phi` or `phi=(hc)/(lamda)-KE_(max)` Also, `KE_(max)=ev_s=0.36eV` `impliesphi=((6.62xx10^(34))xx(3xx10^(8)))/(5893xx10^(-10))-0.36xx1.6xx10^(-19)` `=1.746eV` The threshold frequency is given by `f_0=(phi)/(h)=(2.794xx10^(-19))/(6.62xx10^(-34))=4.22xx10^(14)Hz` |
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| 130. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C. Q. If the mass of the particle is `m=1.0xx10^(-30)`kg and `alpha=6.6nm`, the energy of the particle in its ground state is closest toA. 0.8 meVB. 8 meVC. 80 meVD. 800 meV |
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Answer» `E=(h^2)/(8a^2m)=((6.6xx10^(-34))^2)/(8xx(6.6xx10^(-9))^2xx10^(-30)xx1.6xx10^(-19))` `=8meV` |
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| 131. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)`C Q. The speed of the particle that can take discrete values is proportional toA. `n^((-3)/(2))`B. `n^(-1)`C. `n^((1)/(2))`D. n |
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Answer» `mv=(nh)/(2a)` or `v=(nh)/(2am)impliesvpropn` |
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| 132. |
The radius of the orbit of an electron in a Hydrogen - like atom is `4.5s_(0)` where `a_(0)` is the bohr radius its orbital angular momentum is `(3h)/(2 pi) ` it is given that is is plank constant and R is rydberg constant .The possible wavelength `(s)` , when the atom de- excite , is (are)A. `(9)/(32R)`B. `(9)/(16R)`C. `(9)/(5R)`D. `(4)/(3R)` |
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Answer» Correct Answer - A::C Given data `4.5_(alpha_0)=alpha_0(n^2)/(Z)` `(nh)/(2pi)=(3h)/(2pi)` So `n=3` and `z=2` So possible wavelength are `(1)/(lamda_1)=RZ^2[(1)/(1^2)-(1)/(3^2)]implieslamda_1=(9)/(32R)` `(1)/(lamda_2)=RZ^2[(1)/(1^2)-(1)/(2^2)]implieslamda_2=(1)/(3R)` `(1)/(lamda_3)=RZ^2[(1)/(2^2)-(1)/(3^2)]implieslamda_3=(9)/(5R)` |
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| 133. |
When a particle is restricted to move along x-axis between `x=0` and `x=a`, where `alpha` if of nenometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends `x=0` and `x=a`. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as `E=(p^2)/(2m)`. Thus the energy of the particle can be denoted by a quantum number `n` taking values 1,2,3, ...(`n=1`, called the ground state) corresponding to the number of loops in the standing wave. Use the model described above to answer the following three questions for a particle moving along the line from `x=0` to `x=alpha`. Take `h=6.6xx10^(-34)Js` and `e=1.6xx10^(-19)` C. Q. The allowed energy for the particle for a particular value of n is proportional toA. `alpha^(-2)`B. `alpha^((-3)/(2))`C. `alpha^(-1)`D. `alpha^2` |
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Answer» `a=(nlamda)/(2)implieslamda=(2a)/(n)` `lamda_(de Brogl ie)=(h)/(p)` `(2a)/(n)=(h)/(p)impliesp=(nh)/(2a)` `E=(p^2)/(2m)=(n^2h^2)/(8a^2m)` `impliesEprop(1)/(a^2)` |
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| 134. |
The work function for sodium surface is 2.0eV and that for aluminium surface is 4.2eV. The two metals are illuminated with appropriate radiations so as to cause protoemission. ThenA. the threshold frequency for sodium will be less than that for aluminiumB. the threshold frequency of sodium will be more than that of aluminiumC. both sodium and aluminium will have the same threshold frequencyD. none of the above |
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Answer» Correct Answer - A As work function `W=hv_0`, where `v_0` is the threshold frequency. Greater the work function, greater is the threshold frequency. Therefore, the threshold frequency of sodium will be lesser than that for aluminium. |
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| 135. |
Representing the stopping potential V along y-axis and `((1)/(lamda))` along x-axis for a given photocathode, the curve is a straight line, the slope of which is equal toA. `(e)/(hc)`B. `(hc)/(e)`C. `(ec)/(h)`D. `(e)/(W)` |
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Answer» Correct Answer - B The maximum KE of the photoelectron 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)` Slope of straight line `=(hc)/(e)` Intercept of straight line `=-((W)/(e))` |
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| 136. |
A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum KE of emitted photoelectrons will become.A. `((1)/(16))`th of original valueB. unchangedC. twice the original valueD. four times the original value |
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Answer» Correct Answer - B Maximum KE depends on the frequency of incident radiation, not on intensity. |
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| 137. |
Radiation of wavelength `lambda` in indent on a photocell . The fastest emitted electron has speed `v` if the wavelength is changed to `(3 lambda)/(4)` , then speed of the fastest emitted electron will beA. `v((3)/(4))^((1)/(2))`B. `v((4)/(3))^((1)/(2))`C. less than `v((4)/(3))^((1)/(2))`D. greater than `v((4)/(3))^((1)/(2))` |
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Answer» Correct Answer - D `(1)/(2)mv^2=(4hc)/(3lamda)-W_0` Let the speed of the fastest electron be `V_1` when excitation wavelength is changed to `(3lamda)/(4)`. `(1)/(2)mv_1^2=(4hc)/(3lamda)-W_0` `implies(1)/(2)mv_1^2=(4)/(3)((hc)/(lamda)-W_0)+(W_0)/(3)` `implies(1)/(2)mv_1^2=(4)/(3)((1)/(2)mv^2)+(W_0)/(3)` `impliesv_1^2=(4v^2)/(3)+(2W_0)/(3m)` `v_1gtsqrt((4)/(3))v` |
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| 138. |
Ultraviolet light of wavelength 300nn and intensity `1.0Wm^-2` falls on the surface of a photosensitive material. If one per cent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of 1.0 `cm^2` of the surface is nearlyA. `9.61xx10^14s^-1`B. `4.12xx10^13s^-1`C. `1.51xx10^12s^-1`D. `2.13xx10^11s^-1` |
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Answer» Correct Answer - C `n=(power)/((hc)/(lamda))=(300xx10^-9)/(6.6xx10^-34xx3xx10^8)` `=1.5xx10^18m^-2s^-1=1.5xx10^14cm^-2s^-1` As only 1 percent of photons cause emission of photo-electrons, number of photo electrons is `n_e=1.5xx10^12s^-1` |
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| 139. |
If the short wavelength limit of the continous spectrum coming out of a Coolidge tube is `10 A`, then the de Broglie wavelength of the electrons reaching the target netal in the Coolidge tube is approximatelyA. `0.3A`B. `3A`C. `30A`D. `10A` |
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Answer» Correct Answer - A We have `KE=(p^2)/(2m_e)=(hc)/(lamda_(min))` `P=sqrt((2hcm_e)/(lamda_(min)))` For `lamda_(min)=10A` `lamda_(de Broglie)cong0.3A` |
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| 140. |
The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate . Light source is put on and a saturation photo current is recorded . An electric field is switched on which has a vertically downward direction . Then A. the photocurrent will increaseB. the K.E. of the electrons will increaseC. the stopping potential will decreaseD. the threshold wavelength will increase |
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Answer» Correct Answer - B Electric force on electrons will increase velocity |
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| 141. |
When photons of energy `4.25 eV` strike the surface of metal A, the ejected photoelectrons have maximum kinetic energy `T_(A)` eV and De-broglie wavelength `lambda_(A)`. The maximum energy of photoelectron liberated from another metal B by photon of energy 4.70 eV is `T_(B) = (T_(A) - 1.50) eV` if the de Brogle wavelength of these photoelectrons is `lambda_(B) = 2 lambda_(A) `, thenA. the work function of A is 2.25 eVB. the work funtion of B is 4.20 eVC. `T_A=2.00eV`D. `T_B=2.75eV` |
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Answer» Correct Answer - A::B::C For metal `A:4.25=W_A+T_A` Also, `T_A=(1)/(2)mv_A^2=(1)/(2)(m^2v_A^2)/(m)=(p_A^2)/(2m)(h^2)/(2mlamda_A^2)` For metal `B:4.7=(T_A-1.5)+W_B` Also, `T_B=(h^2)/(2mlamda_B^2)xx(2mlamda_A^2)/(h^2)=(lamda_A^2)/(lamda_B^2)` `implies(T_A-1.5)/(T_A)=(lamda_A^2)/(2lamda_A^2)=(lamda_A^2)/(4lamda_A^2)=(1)/(4)` `implies4T_A-6=T_AimpliesT_A=2eV` |
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| 142. |
An `alpha` particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are `lambda_a` and `lambda_p` respectively. The ratio `(lambda_p)/(lambda_a)`, to the nearest integer, is. |
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Answer» `(1)/(2)mv^2=qV` `lamda=(h)/(mv)` `lamda=sqrt8=3` |
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