Saved Bookmarks
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.
| 151. |
The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted, when it finally moves to the ground state? |
|
Answer» for third excited state, n=4 and for ground state n=1. Hence, the possible transitions are: `n_i=4` to `n_f=3,2,1` `n_i=3` to `n_f=2,1` `n_i=2` to `n_f=1` Hence, total number of transitions=6 |
|
| 152. |
During beta decay of a nucleus, how does the neutrons to proton ration change? |
|
Answer» Let us consider `beta`-decay of `._83Bi^(210)` `._83Bi^(210) to ._84Po^(210)+ ._-1e^0` Neutron to proton ratio before `beta`-decay `=(210-83)/83=127/83`, and Neutron to proton ratio after `beta`-decay `=(210-84)/84=126/84`. which is less than `(127)/(83)`. Hence, the neutron to proton ratio decrease in `beta`-decay. |
|
| 153. |
The mass number of a nucleus isA. always less than its atomic numberB. always more than its atomic numberC. sometimes equal to its atomic numberD. sometimes more than and sometimes equal to its atomic number |
|
Answer» Correct Answer - C::D A and Z are sometimes equal e.g. for hydrogen nucleus `(._(1)H^(1)), Z=A=1`, for other nuclei (e.g. `._(6)C^(13), A=13, Z=6`) so AgtZ. |
|
| 154. |
Consider a hydrogen atom with its electron in the `n^(th)` orbital An electomagnetic radiation of wavelength `90 nm` is used to ionize the atom . If the kinetic energy of the ejected electron is `10.4 eV` , then the value of `n` is ( hc = 1242 eVnm)A. `1`B. `2`C. `3`D. `4` |
|
Answer» Correct Answer - B Here, `lambda=90nm, K=10.4 eV` As `E_(n)=-13.6/(n^(2))eV` Now `(hc)/lambda-(13.6 eV)/(n^(2))=10.4eV` `(1242eV nm)/(90 nm)-(13.6eV)/(n^(2))=10.4` or `4.14/3-13.6/(n^(2)) or 13.8-10.4=13.6/(n^(2))` or `3.4=13.6/(n^(2)) or n^(2)=13.6/3.4 or n^(2)=4` to `n=2` |
|
| 155. |
An isotope `._(92)U^(238)` decays successively to form `._(90)Th^(234),._(91)Pa^(234),._(90)Th^(234),._(90)Th^(230)` and `._(88)Ra^(226)`. What are the radiations emitted in these five steps? |
| Answer» Correct Answer - `alpha,beta, beta, alpha,alpha` | |
| 156. |
A nucleus of uranium decays at rest into nuclei of thorium and helium. Then :A. The helium nucleus has less kinetic energy than the thorium nucleusB. The helium nucleus has more kinetic energy than the thorium nucleusC. The helium nucleus has less momentum than the thorium nucleusD. The helium nucleus has more momentum than the thorium nucleus |
|
Answer» Correct Answer - B `._(92)U^(238)to._(90)Th^(234)+._(2)He^(4)+"Energy"` KE of thorium `=(p^(2))/(2m_(Th))` and KE of helium `=(p^(2))/(2m_(He))` Since `m_(He) lt m_(Th)` so `KE_(He) gt KE_(Th)` |
|
| 157. |
In the final Uranium radioactive series the initial nucleus is `U_(92)^(238) ` and the final nucleus is `Pb_(82)^(206)` . When Uranium neucleus decays to lead , the number of a - particle is …….. And the number of `beta` - particles emited is …… |
|
Answer» Correct Answer - `8alpha and 6 beta` Let x be the number of alpha particles and y be the no. of `beta` particles emitted in the disintegration process of `._(92)U^(238)` to `._(82)Pb^(206)`. We may write `._(92)U^(238) to._(82)Pb^(206)+(._(2)He^(4))+y(._(-1)e^(0))` form law of conservation of mass number `238=206+4x+y(0)=206+4x` `=4x=238-206=32, x=8` form law of conservation of mass number, `92=82+2x-y=-10+2x` `y=-10+2(8)=6` Hence `8alpha` particles and `6 beta` particles are emitted. |
|
| 158. |
The atomic mass of uranium `._(92)^(238)U` is `23.058 u`, that of throium` ._(90)^(234)Th` is `234.0436 u` and that of an alpha particle `._2^4He` is `4.006 u`, Determine the energy released when `alpha-decay` converts`._(92)^(238)U` into `._(92)^(238) U`. int `._(90)^(234)Th`. |
|
Answer» The given nuclear reaction is `._92U^(238) to ._90Th^(234)+._2He^4` Mass defect, `Deltam=238.05079-234.04363-4.000260` `=0.00456 u` `:.` Energy released `=0.00456xx931.5MeV` `=4.25MeV` |
|
| 159. |
A uranium nucleus `U-238` of atomic number`92` emits two `alpha`- particles and two `beta`-particles and trasnforms into a thorium nucleus. What is the mass number and atomic number of the thorium nucleus so produced? |
|
Answer» Alpha particle is `._2He^4` and Beta particles is `._(-1)e^0`. `:.` on emission of `2alpha` and `2beta` particles, mass no. of Thorium=`235-2xx4=227` Charge no. of Thorium `=92-2xx2-2(-1)=90` |
|
| 160. |
Write nuclear reaction equation for (i) `alpha` particle of `._88Ra^(226)` (ii) `alpha` decay of `._94Pu^(242)` (ii) `beta^(-)` decay of `._15P^(32)` (iv) `beta^(-)` decay of `._83Bi^(210)` (v) `beta^(+)` decay of `._6C^(11)` (vi) `Beta^(+)` decay of `._43Tc^(97)` (vii) Electron capture of `._54Xe^(120)`. |
|
Answer» (i) `._88Ra^(226) to._86Rn^(222)+._2He^4` (ii) `._94Pu^(242) to ._92U^(238)+._2He^4` (iii) `._15P^(32)to._16S^(32)+e^(-)+barv` (iv) `._83Bi^(210)to._84X^(210)+e^(-)+barv` (v) `._6C^(11)to._5B^(11)+e^(+)+v` (vi)`._43Tc^(97) to._42X^(97)+e^(+)+v` (vii) `._54Xe^(120)+._(-1)e^0to ._(53)X^(120)` |
|
| 161. |
(a) Write symbolically the process expressing the `beta^(+)` decay of `._11Na^(22)`. Also write the basic nuclear process underlying this decay. (b) Is the nucleus formed in the decay of the nucleus `._11Na^(22)`, an isotope or isobar? |
|
Answer» (a) `underset (underset(("parent nucleus"))darr )(._11Na^(22)) tounderset (underset(("Daughter nucleus"))darr )(._10Ne^(22))+underset (underset((beta^(+)))darr )(._(+1)e^0)+underset (underset("Energy")darr )(Q)` The basic nuclear process is decay by emission of `beta^(+)` particle. (b) The nucleus formed in the decay of `._11Na^(22)` is obviously an isobar having same mass number but different charge number. |
|
| 162. |
The radii of stationary orbits are in the ratio............ . |
| Answer» Correct Answer - `1^(2):2^(2):3^(2)` and so on, | |
| 163. |
Assertion: Balmer series lies in visible region of electromagnetic spectrum. Reason: Balmer means visible, hence series lies in visible region.A. If both, Assertion and Reason are true and the Reason is the correct explanation of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explanation of the Assertion.C. If Assertion is true but the Reason is false.D. If both Assertion and Reason are false. |
|
Answer» Correct Answer - C Assertion is true, but the reason is false. for Balmer series, `1/lambda=R[1/(2^(2))-1/(n^(2))]`, where n=3,4,5...... When we put n=3,4,5....and `R=10^(-7)m^(-1)` in the given formula, the values of `lambda` calculated lie between 4000Å to 8000Å, which is visible region. |
|
| 164. |
According to Bohr, angular momentum...............in hydrogen atom is........... . |
| Answer» Correct Answer - of revolution electron; quantised, | |
| 165. |
Lyman series lies in............region and Balmer series lies in.......... region of hydrogen spectrum. |
|
Answer» Correct Answer - ultraviolet; visible, Visible is the right answer
|
|
| 166. |
Calculate the maximum energy that a `beta` particle can have in the following decay: `._(8)O^(19) to ._(9)F^(19)+._(-1)e^(0)+barV` Given, `m(._(8)O^(19))=19.003576u, m(._(9)F^(19))=18.998403u, m(._(-1)e^(0))=0.000549u` |
| Answer» Correct Answer - `4.3MeV` | |
| 167. |
What is the difference between a beta particle and electron? |
| Answer» An electron and a `beta`-particle are essentially the same. We can say that a `beta`-particle is an electron of nuclear origin. | |
| 168. |
What is the relation between decay constant and half life of a radio active element? |
| Answer» Correct Answer - `T=0.693//lambda` | |
| 169. |
The decay constant for a radio nuclide, has a value of `1.386 day^(-1)`. After how much time will a given sample of this nuclide get reduced to only 6.25% of its present number ? |
|
Answer» Correct Answer - `2days` Here, `lambda=1.386 day^(-1), t=?` `N/(N_(0))=6.25%` Half life `T=0.693/lambda=0.693/1.386=1/2=0.5 day` `N/(N_(0))=(1/2)^(n)=(6.25/100)=1/16=(1/2)^(4)` `:. n=4=t/T, t=4T=4xx0.5=2 days` |
|
| 170. |
One electron volt is the...........when accelerated through a............. . |
| Answer» Correct Answer - energy acquired by an electron; potential difference of 1 volt. | |
| 171. |
Calculate the energy equivalent of 1 a.m.u. in `MeV` |
|
Answer» form `E=mc^2`, when `m=1 a.m.u. =1.66xx10^(-27)kg` `E=(1.66xx10^(-27))(3xx10^8)^2 "jouls"` `=(1.66xx9xx10^(-11))/(1.6xx10^(-13)) MeV=933.75MeV` |
|
| 172. |
Calculate the energy equivalent of 1g of substance. |
|
Answer» Here, `m=1 g=10^-3 kg, E=?` form `E=mc^2=10^-3(3xx10^8)^2=9xx10^(13)J` |
|
| 173. |
Calculate the binding energy per nucleon of `._20Ca^(40)` nucleus. Given m `._20Ca^(40)=39.962589u` , `M_(p) =1.007825u` and `M_(n)=1.008665 u` and Take `1u=931MeV`. |
|
Answer» In a nucleus of `._20Ca^(40)`, number of protons=20, number of neutrons `=40-20=20` Total mass of 20 protons and 20 neutrons `=20m_p+20m_n=20(m_p+m_n)` `=20(1.007825+1.008665)=40.3298u` Mass defect, `Delta m=40.3298-39.962589` `=0.367211 u` Total `B.E=0.367211xx931MeV` `=341.873441 MeV` `BE//"nucleon"=341.873441/40=8.547MeV//N` |
|
| 174. |
Find the minimum energy required to separated a neutron form `._18Ar^(40)`. Given `M_n=1.008665 u` , mass of `._18Ar^(40)=39.962383u` and mass of `._18Ar^(39)=38.964314u`. |
|
Answer» The nuclear reaction involved is `._18Ar^(40)` to `._18Ar^(39)+._0n^1` Mass defect `=39.962383-(38.964314+1.008665)` `Deltam=0.010596u` `:.` Minimum separation energy required `=Delta mxx931MeV` `=0.010596xx931=9.86MeV` |
|
| 175. |
Write the SI unit of activity of a radioactive nuclide. |
|
Answer» The SI unit of activity is becquerel 1 bacquerel =1Bq=1decay/sec. |
|
| 176. |
The decay constant for a given radioactive sample is 0.3465 `day^-1`. What percentage of this sample will get decayed in a period of 4 days? |
|
Answer» Here, `lambda=0.3465 day^-1, t=4days`. Half life, `T=0.693/lambda=0.693/0.3465=2days` No. of half lives, `n=t/T=4/2=2` `:. N/(N_0)=(1/2)^n=(1/2)^2=1/4=25%` Percentage of sample that will get decayed =100-25=75% |
|
| 177. |
The mean life of a radioactive sample is `T_(m)`. What is the time in which `50%` of the sample woulf get decayed ? |
|
Answer» Time in which `50%` of the sample will get decayed is half life `T=0.693/lambda=0.693tau=0.693 T_m`. |
|
| 178. |
A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life of one species is `tau` and that of the other is `5 tau`. The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figure best represents the form of this plot? (a), (b), (c), (d) A. B. C. D. |
|
Answer» Correct Answer - D Both the species of atoms follow exponential decay law, whatever be their life time. Therefore, option (d) is correct. |
|
| 179. |
The half life of Radon is 3.8 days. Calculate how much of 15 milligram of Redon will remain after 38 days. |
|
Answer» Here, `T=3.8days, t=38 days, N_0=15 "milligram", N=?` No. of half lives in 38 days is `n=(38)/(3.8)=10` We know, `N=N_0(1/2)^n :. N=15(1/2)^(10)=0.014 mg` |
|
| 180. |
A radio active substance has a half life period of 30 days. Calculate (i) time taken for `3//4` of original number of atoms to disintegrate, (ii) time taken for `1//8` of the original number of atoms to remain unchanged. |
|
Answer» Here, T=30days,t=? No. of atoms disitingrated =`(3//4)N_0` No. of atoms left after time t, `N=N_0-3/4N_0=1/4N_0` No. half lives in time t days, `n=t/T=t/30` No. of nuclei left after n half lives is given by `N=N_0(1/2)^n` `:. (N_0)/4=N_0(1/2)^(t//30) or (2)^(t//30)=4=2^2` or`t/30=2 or t=60 days` (ii) `t=? ,N=N_0//8` As `N=N_0(1/2)^n :. (N_0)/8=N_0(1/2)^(t//30)` or `(2)^(t//30)=8=(2)^3 or t/30 =3` or `t=90 days ` |
|
| 181. |
The bombardment of Lithium with protons gives rise to the following reaction: `._3Li^7+._1H^1 to .2_He^4+.2_He^4+Q` The atomic masses of lithium, hydrogen and helium are : `7.016u, 1.008u` and `4.004u` resp. Find the initial energy of each of `alpha` particle. Take `1a.m.u=931MeV`. |
|
Answer» Here, we are going atomic masses. In the given nuclear reaction, masses of 4 electrons before and after reaction cancel out. Therefore, we can use these masses as corresponding nuclear masses. `Q=Deltamxx931MeV` `=[m(Li)+m(H)-2m(He)]xx931MeV` `=[7.016+1.008-2xx4.004]xx931` `=0.016xx931=14.896MeV` This energy is shared equally between two alpha particles. `:.` Energy of each `alpha` particle `=14.896/2=7.448MeV` |
|
| 182. |
(a) Two stable isotope of `._3Li^6` and `._3Li^7` have respective abundances of `7.5%` and `92.5%`. These isotopes have masses 6.01512 and 7.01600 u respectively. Find the atomic weight of lithium. (b) Boron has two stable isotopes `._5B^(10)` and `._5B^(11)`. Their respective masses are 10.01294 u and 11.00931u, and the atomic weight of boron is 10.81u. Find the abundaces of `._5B^(10)` and `._5B^(11)`. |
|
Answer» (a) Atomic weight =weigthed average of the isotopes `=(6.01512xx7.5+7.01600xx92.5)/((7.5+92.5))=(45.1134+648.98)/100=6.941u` (b) Let relative abundace of `._5B^(10)` be x% `:.` Relative abundance of `._5B^(11)=(100-x)%` Proceeding as abve, `10.811=(10.01294x+11.00931xx(100-x))/100` `x=19.9% and (100-x)=80.1%` |
|
| 183. |
The groud state energy of hydrogen atom is `-13.6 eV`. When its electron is in first excited state, its exciation energy is |
|
Answer» Energy in ground state `E_1=-13.6eV`. Energy is first excited state, `E_2=(-13.6)/(2^2)=-3.4eV.` `:.` Required energy =`E_2-E_1` `=-3.4-(-13.6)=10.2eV` |
|
| 184. |
Isotones are the nuclides which contain............ . |
| Answer» Correct Answer - same number of neutrons. | |
| 185. |
The ground state energy of hydrogen atom is -13.6 eV (i) What are the potential energy and K.E. of electron is 3rd excited state? (ii) If the electron jumped to the ground state form the third excited state, calculate the frequency of photon emitted. |
|
Answer» Correct Answer - `(i) -1.7eV; 0.85eV ; (ii) 3xx10^(15)Hz` Here, `E_(1)=-13.6eV` for third excited state, n=4 `:. E_(4)=(-13.6)/(4^(2))=-0.85eV` `:. K.E.=-E_(4)=0.85eV` `P.E. =-2 (K.E.)=-2(0.85)eV=-1.70eV` Energy emitted, `DeltaE=E_(4)-E_(1)` `hv=-0.85-(-13.6)eV=12.75eV` `v=(12.75xx1.6xx10^(-19))/(6.6xx10^(-34))=3xx10^(15) Hz` |
|
| 186. |
Samples of two radioactive nuclides A and B are taken. `lambda_(A) and lambda_(B)` are the disintegration constants of A and B respectively. In which of the following cases, the two samples can simultaenously have the same decay rate at any time?A. Initial rate of decay of A is twice the initial rate of decay of B and `lambda_(A)=lambda_(B)`B. Initial rate of decay of A is twice the initial rate of decay of B and `lambda_(A)gtlambda_(B)`C. Initial rate of decay of B is twice the initial rate of decay of A and `lambda_(A)gtlambda_(B)`D. Initial rate of decay of B is same as the rate of decay of A at t=2h and `lambda_(B)=lambda_(A)` |
|
Answer» Correct Answer - B::D The two samples of two radioactive nuclides A and B can simultaneously have the same decay rate at any time if initial rate of of decay A is twice the initial rate of decay of B and `lambda_(A)gtlambda_(B)`. Also, when initial rate of decay of B is same as rate of decay of A at t=2h and `lambda_(B) ltlambda_(B)`. The choice (b) and (d) is correct. |
|
| 187. |
A `12.5 eV` electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted. |
| Answer» In ground state, energy of gaseous hydrogen at room temperature `=-13.6eV.` When it is bombarded with 12.5 eV electron beam, the energy becomes `-13.6+12.5=-1.1eV`. The electron would jump form n=1 to n=3, where `E_3=-13.6/(3^2)=-1.5eV`. On de-excitation the electron may jump form n=3 to n=2 giving rise to Balmer series. It may also jump form n=3 to n=2 giving rise to Balmer series. It may also jump form n=3 to n=1, giving rise to Lyman series, | |
| 188. |
The wavelength of some of the spectral lines obtained in hydrogen spectrum are `1216 A^(@),6463A^(@) and 9546 A^(@)`. Which one of these wavelengths belongs to the Paschen series? |
| Answer» `lambda=9546Å` belongs to the Paschen series | |
| 189. |
The energy of a hydrogen atom in the ground state is `-13.6 eV`. The eneergy of a `He^(+)` ion in the first excited state will beA. `-13.6eV`B. `-27.2 eV`C. `-54.4eV`D. `-6.8 eV` |
|
Answer» Correct Answer - A `E_(He^(+))=-(Z^(2))/(n^(2))xx13.6eV` for `He^(+)` ion, Z=2, and for first excited state, n=3. `:. E_(He^(+))=-(2^(2))/(2^(2))xx13.6eV=-13.6 eV` |
|
| 190. |
The wavelength of the first spectral line in the Balmer series of hydrogen atom is `6561 A^(@)` . The wavelength of the second spectral line in the Balmer series of singly - ionized helium atom isA. `1215Å`B. `1640Å`C. `2430Å`D. `4687Å` |
|
Answer» Correct Answer - A The wavelength `(lambda)` of a spectral line in the Balmer series is given by `1/lambda=RZ^(2)[1/(2^(2))-1/(n^(2))]` where n=3,4,5,6..... In case of hydrogen atom, for first spectral line in the Balmer series, Z=1, n=3 , `lambda=lambda_(H)` (say) `:. 1/(lambda_(H))=Rxx(1)^(2)[1/(2^(2))-1/(3^(2))]=Rxx5/36.......(i)` In case of singly -ionized helium atom, for second spectral line in the Balmer series, Z=2 , n=4, `lambda=lambda_(He^(+))` `1/(lambda_(He^(+)))=Rxx(1)^(2)[1/(2^(2))-1/(4^(2))]= 4Rxx3/16.......(ii)` Dividing (i) by (ii), we get `(lambda_(He^(+)))/(lambda_(H))=(5//36)/(4xx3//16)=5/27` `lambda_(H^(+))= lambda_(H) xx5/27=(6561Å)xx5/27=1215Å` |
|
| 191. |
`M_(x) and M_(y)` denote the atomic masses of the parent and the dougther nuclei respectively in a radioactive decay. The Q - value for a `beta-` decay is `Q_(1)` and that for a `beta^(+)` decay is `Q_(2)`. If `m_(e)` denotes the mass of an electrons, then which of the following statements is correct?A. `Q_(1)=(M_(x)-M_(y))c^(2) and Q_(2)=(M_(x)-M_(y)-2m_(e))c^(2)`B. `Q_(1)=(M_(x)-M_(y))c^(2) and Q_(2)=(M_(x)-M_(y))c^(2)`C. `Q_(1)=(M_(x)-M_(y)-2m_(e))c^(2) and Q_(2)=(M_(x)-M_(y)+2m_(e))c^(2)`D. `Q_(1)=(M_(x)-M_(y)+2m_(e))c^(2) and Q_(2)=(M_(x)-M_(y)+2m_(e))c^(2)` |
|
Answer» Correct Answer - A `beta^(-)` decay is represented as `._(Z)X^(A)to._(Z+1)A^(Y)+._(-1)e^(0)+barv+Q_(1)` `:. Q_(1)=[m_(N)(._(Z)X^(A))-m_(N)(._(Z+1)Y^(A))-m_(e)]c^(2)=[m_(N)(._(Z)X^(A))+Zm_(e)-m_(N)(._(Z+1)Y^(A))-(Z+1)m_(e)]c^(2)` `=[m(._(Z)X^(A))-m(._(Z+1)Y^(A))]c^(2)=(M_(x)-M_(y))c^(2)` `beta^(+)` decay is represented as `._(Z)X^(A)=._(Z-1)Y^(A)+._(1)e^(0)+v+Q_(2)` `:. Q_(2)=[m_(N)(._(Z)X^(A))-m_(N)(._(Z-1)Y^(A))-m_(e)]c^(2)=[m_(N)(._(Z)X^(A))+Zm_(e)-m_(N)(._(Z-1)Y^(A))-(Z-1)m_(e)-2m_(e)] c^(2)` `=[m(._(Z)X^(A))-m(._(Z-1)Y^(A))-2m_(e) ]c^(2)=(M_(x)-M_(y)-2m_(e))c^(2)` `:.` Choice (a) is correct. |
|
| 192. |
Tritium is an isotope of hydrogen whose nucleus Triton contains 2 neutrons and 1 proton. Free neutrons decay into `p+bare+barv`. If one of the neutrons in Triton decays, it would transform into `He^(3)` nucleus. This does not happen. This is becauseA. Triton energy is less than that of a `He^(3)` nucleusB. the electron created in the beta decay process can not remain in the nucleusC. both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a `He^(3)` nucleusD. because free neutrons decay due to external perturbations which is absent in a triton nucleus |
|
Answer» Correct Answer - A Tritium `to ._(1)H^(3)`. The nucleus contains 1 proton and 2 neutrons. If one neutrons decays `ntop+e^(-)+barv`, the nucleus may have 2 protons and one neutron, i.e., tritium will transform into `._(2)He^(3)` (with 2 protons and one neutron). But this does not happen because triton energy is less than that of `._(2)He^(3)` nucleus , i.e., transformation is not allowed energetically. Choice (a) is correct. |
|
| 193. |
A radioactive sample contains 2.2 mg of pure `._6C^(11)` , which has a half life period of 1224s. Calculate (i) number of atoms present initially. (ii) the activity when `5 mu g` of the sample will be left. |
|
Answer» Correct Answer - `1.2xx10^(20); 1.55xx10^(14)` disintegration/sec Here, number atoms present in 11 gram of sampel `=6.023xx10^(23)` `:.` No. of atoms present in 2.2 mg of sample initially `N_(0)=(6.023xx10^(23)xx2.2xx10^(-3))/11=1.2xx10^(20)` No. of atoms present in `5 mug` `N=(6.023xx10^(23)xx5xx10^(-6))/11=2.74xx10^(17)` Activity of the sample `A=lambdaN=0.693/TN=(0.693xx2.74xx10^(17))/1224` `A=1.55xx10^(14)` disintegration/sec. |
|
| 194. |
Calculate binding energy per nucleon of `._83Bi^(209)`. Given that m `(._83Bi^(209))=208.980388am u` `m("neutron") = 1.008665 am u` `m ("proton") = 1.007825 am u` |
|
Answer» In `.83Bi^(209)`, number of proton=83 number of neutrons `=209-83=126` Mass defect, `Delta m=83xxm_p+126xxm_n-M(Bi)` `=83xx1.007825+126xx1.008665-208.980388` `=83.6494475xx127.091790-208.980388` `=1.760877 am u` `B.E. //"nucleon"=(1.760877xx931)/209` `=7.85MeV//N` |
|
| 195. |
Obtain the binding energy of the nuclei `._26Fe^(56)` and `._83Bi^(209)` in units of MeV form the following data: `m(._26Fe^(56))=55.934939a.m.u.` , `m=(._83Bi^(209))=208.980388 a m u`. Which nucleus has greater binding energy per nucleon? Take `1a.m.u 931.5MeV` |
|
Answer» (i) `._(26)F^(56)` nucleus contains 26 protons and `(56-26)=30` neutrons Mass of 26 protons `=26xx1.007825=26.20345 a.m.u` Mass of 30 neutrons `=30xx1.008665=30.25995 a.m.u` Total mass of 56 nuleons `=56.46340 a.m.u`. mass of `._26Fe^(56)` nucleus `=55.934939` `:.` Mass defect `Deltam=56.46340-55.934939=0.528461 a.m.u.` Total binding energy `=0.528461xx931.5MeV=-492.26 MeV` Av. B.E. per nucleon=492.26/56=8.790MeV (ii) `._83Bi^(209)` nucleus 83 protons and `(209-83)=126` neutrons. Mass of 83 protons =`83xx1.007825=83.649475 a.m.u.` Mass of 126 neutrons `=126xx1.008665=127.091790 a.m.u.` Total mass of nucleons `=210.741260 a.m.u`. Mass of `._83Bi^(209)` nucleus `=208.980388 a.m.u`. Mass defect, `Deltam=210.741260-208.980388=1.760872` Total B.E. `=1.760872xx931.5MeV=1640.26 MeV.` Av. B.E. per nucleon =`1640.26/209=7.848 MeV` Hence `._26F^(56)` has greater B.E. per nucleon than `._83Bi^(209)` |
|
| 196. |
In a head on collision between an alpha particle and gold nucleus (Z=79), the distance of closest approach is 39.5 fermi. Calculate the energy of `alpha`-particle. |
|
Answer» Here, Z=79, `r_0=39.5fermi=39.5xx10^(-15)m , E=?` form `E=1/2mv^2=1/(4pi in_0)((Ze)(2e))/(r_0)` `E=(9xx10^9xx79xx2(1.6xx10^(-19))^2)/(39.5xx10^(-15))` `=(18xx79xx2.56)/(39.5xx10^(-14))J` `=(9.216xx10^(-13))/(1.6xx10^(-13))=5.76MeV` |
|
| 197. |
Express 1 joule in eV. Taking 1a.m.u. `=931 MeV`, calculate the mass of `._6C^12`. |
|
Answer» As `1eV=1.6xx10^(-19)J` `:. 1J=1/(1.6xx10^(-19))eV=6.25xx10^(18)eV` Mass of C-12 atom `12am u=12xx931MeV` `=12xx931xx1.6xx10^(-13)J` `=(12xx931xx1.6xx10^(-13))/((3xx10^8)^2)kg` `m=1.986xx10^(-26)kg` |
|
| 198. |
Suppose, we think of fission of a `._26Fe^(56)` nucleus into two equal fragments `._13Al^(28)`. Is the fission energetically possible? Argue by working out Q of the process. Given `m(._26Fe^(56))=55.93494u, m(._13Al^(28))=27.98191 u.` |
|
Answer» `Q=[m(._26Fe^(56))-2m(._13Al^(28))]xx931.5MeV=[55.93494-2xx27.9819]xx931.5MeV` `Q=-0.02886xx931.5MeV=-26.88 MeV`, which is negative. The fission is not possible energetically. |
|
| 199. |
The radius of `._3Al^(27)` nucleus is 5 fermi. Find the radius of `._52Te^(125)` nucleus. |
|
Answer» Here, `A_1=27, R_1=6 fermi` `A_2=125,R_2=?` As `(R_2)/(R_1)=((A_2)/(A_1))^(1//3)=(125/27)^(1//3)=5/3` `R_2=5/3R_1=5/3xx6=10fermi` |
|
| 200. |
Let `E_(n)=(-1)/(8epsilon_(0)^(2)) (me^(4))/(n^(2)h^(2))` be the energy of the `n^(th)` level of H-atom. If all the H-atom are in the ground state and radiation of frequency `(E_(2)-E_(1))//h` falls on it,A. it will not be absorbed at allB. some of atoms will move to the first excited stateC. all atoms will be excited to the n=2 stateD. no atoms will make a transition to the n=3 state |
|
Answer» Correct Answer - B::D Here, `E_(n)=(me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))` Is the energy of nth level of hydrogen atom. If all the H-atom are in ground state, (n=1), then the radiation of frequency `(E_(2)-E_(1))//h` falling on it may not be absorbed by some of the atoms and move them to the first excited state (n=2). All atoms may not be excited to n=2 state. Further, as `(E_(2)-E_(1))//h` is sufficient only to take the atom form n=1 state to n=2 state, no atoms shall make a transition to n=3 state. Choices (b) and (d) are correct. |
|