InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 1. |
The radioactive nuclei are unstable and emit alpha `(._(2)He^(4))` , beta `(._(-1)e^(0))` and gamma `(._(0)gamma^(0))` radiations to acheive states of greater stability. Rutherford and Soddy suggested the following rules governing the radioactive decay, (i) The algebric sum of charges (atomic numbers) before and after disintegration must be the same. (ii) The sum of mass numbers before and after disintegration must also be the same. In beta decayA. A remains unaffected, Z increases by 1.B. A is unaffected, Z decreases by 1C. A increases by 1, Z is unaffactedD. A decreases by 1, Z is unaffected. |
| Answer» As a beta particle carries unit negative charge and has negligible mass, therefore, in beta decay, A remains unaffected and Z increases by 1. | |
| 2. |
The radioactive nuclei are unstable and emit alpha `(._(2)He^(4))` , beta `(._(-1)e^(0))` and gamma `(._(0)gamma^(0))` radiations to acheive states of greater stability. Rutherford and Soddy suggested the following rules governing the radioactive decay, (i) The algebric sum of charges (atomic numbers) before and after disintegration must be the same. (ii) The sum of mass numbers before and after disintegration must also be the same. In alpha decayA. mass number A decrease by 4 and atomic number Z increases by 2.B. A decreases by 4 and Z decreases by 2C. A increases by 4 and Z decreases by 2D. A increases by 4 and Z increases by 2. |
| Answer» As mass of an `alpha`-particle is 4 units and it carries 2 units of positive charge, therefore, in alpha decay, A decreases by 4 and Z decreases by 2 | |
| 3. |
Out of alpha, beta and gamma radiations, which are affected by electric field and magnetic field? |
| Answer» Both, alpha and beta radiations are affected by both, electric field and magnetic field. | |
| 4. |
The mass of a `a_3^7 Li` nucleus is `0.042 u` less than the sum of the masses of all its nucleons. The binding energy per nucleon of `._3^7 Li` nucleus is nearly. |
|
Answer» Correct Answer - `5.588 MeV//N` Here, `Deltam=0.042a.m.u` BE=`Deltamxx931.5MeV` `=0.042xx931.5MeV=39.123MeV` BE/nucleon `=39.123/7=5.588MeV//N` |
|
| 5. |
A hydrogen atom initially in the ground level absorbs a photon, Which excites it to n=4 level. Determine the wavelength and frequency of photon. |
|
Answer» For ground state, `n_1=1 and n_2=4`. Energy of photon absorbed, `E=E_2-E_1=(-13.6)/(n_2^2)-((-13.6)/(n_1^2))eV=13.6(1/(n_1^2)-1/(n_2^2))=13.6(1/(1^2)-1/(4^2))eV` `=13.6xx15/16eV=(13.6xx15)/16 xx1.6xx10^(-19)J=2.04xx10^(-18)"joule"` form `E=(hc)/lambda, lambda=(hc)/E=(6.6xx10^(-34)xx3xx10^8)/(2.04xx10^(-18))=9.7xx10^(-8)m, v=c/lambda=(3xx10^8)/(9.7xx10^(-8))=3.1xx10^(15)Hz` |
|
| 6. |
Energy of an electron in an excited hydrogen atom is `-3.4eV`. Its angualr momentum will be: `h = 6.626 xx 10^(-34) J-s`. |
|
Answer» Correct Answer - `2.11xx10^(-34)Js` `E=(-13.6)/(n^(2))=-3.4eV :. n=2` `:.` Angular momentum of electron `=(nh)/(2pi)=2h//2pi=h/pi=(6.626xx10^(-34))/3.14` `2.11xx10^(-34)J-s` |
|
| 7. |
Nuclei with magic no. of proton Z=2,8,20,28,50,52 and magic no. of neutrons N=2,8,20,28,50,82 and 126 are found to be stable. (i) Verify this by calculating the proton separation energy `S_(p)` for `.^(120)Sn (Z=50)` and `.^(121)Sb =(Z=51)`. The proton separation energy for a nuclide is the minimum energy required to separated the least tightly bound proton form a nucleus of that nuclide. It is given by `S_(p)=(M_(z-1,N)+M_(H)-M_(Z,N))c^(2)`. given `.^(119)Sn =118.9058u, .^(120) Sn =119.902199u, .^(121)Sb=120.903824u, .^(1)H=1.0078252u` (ii) what does the existence of magic number indicate? |
|
Answer» (i) The proton separation energy is given by `S_(p)=(M_(Z-1,N)+M_(H)-M_(Z,N))c^(2)......(i)` for sn, `M_(Z-1,N)=118.9058u, M_(H)=1.0078252u` and `M_(Z,N)=119.902199u` `:.` form (i), `S_(pSn)=(118.9058+1.0078252-119.902199)c^(2)=0.0114362c^(2)` Similarly, `S_(pSb)=(M_(50,70)+M_(H)-M_(51,70)) c^(2)=(119.902199+1.0078252-120.903824)c^(2)` `=0.0059912c^(2)` Since `S_(pSn)gtS_(pSb)`, therefore, `S_(n)` nucleus is more stable than Sb nucleus, which was to be proved. (ii) The existance of magic numbers indicates the shell structure of nucleus similar to the shell structure of an atom. It also accounts for the peaks in Binding energy/nucleon curve. |
|
| 8. |
What is the difference between a photon and a neutrino? |
|
Answer» A photon is one quantum of electromagnetic radiation. It has zero rest mass, zero charge, zero spin and no antiparticles. Its energy `E=hv` depends on its frequency. A neutrino is an elementary particle that accompanies `beta`-decay. It has zero rest mass, zero charge and spin=`1/2(h//2pi)`. If has an antipartical called antineutrino. It can have any energy form zero to a value permitted by nuclear rection. Its nature is non electromagnetic. |
|
| 9. |
Before the neutrino hypothesis, the beta process was thought to be the transition, `n rarr p+bare`. If the was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range. |
|
Answer» We are supporting that `beta` decay is due to the transition `n rarr p+e^(-)` Suppose before beta decay, neutron is at rest. `:. p_(n)=0 and E_(n)=m_(n)c^(2)` After `beta` decay, form conservation of linear momentum, we have `vecp_(n)=vecp_(p)+vecp_(e)=0 :. |vecp_(p)|=|vecp_(e)|=p`, say. `E_(p) =(m_(p)^(2) c^(4)+p_(p)^(2) c^(2))^(1//2) and E_(e)=(m_(e)^(2) c^(4)+p_(e)^(2) c^(2))^(1//2)=(m_(e)^(2) c^(4)+p_(p)^(2) c^(2))^(1//2)` form conservation of energy, `E_(p)+E_(e)=E_(n)` i.e., `(m_(p)^(2) c^(4)+p_(p)^(2) c^(2))^(1//2)+(m_(e)^(2) c^(4)+p_(e)^(2) c^(2))^(1//2)=m_(n)c^(2).......(i)` Now, `m_(p)c^(2)=936MeV , m_(n)c^(2)=938MeV` and `m_(e)c^(2)=0.51MeV`. As energy difference between `m_(p)c^(2)` and `m_(n)c^(2)` is small, therefore pc will be small. `:. pclt lt m_(p)c^(2)`. However, pc may be greater than `m_(e)c^(2)`. therefore, form (i) `m_(p)c^(2)+(p^(2)c^(2))/(2m_(p)^(2)c^(4))~=m_(n)c^(2)-pc` To first order of approx. `pc~=m_(n)c^(2)-m_(p)c^(2)=938MeV-936MeV=2MeV` This gives us the momentum of proton or neutron. `:. E_(p)=(m_(p)^(2) c^(4)+p^(2) c^(2))^(1//2)=sqrt((936)^(2)+2^(2)) ~=936MeV` and `E_(e)=(m_(e)^(2) c^(4)+p^(2) c^(2))^(1//2)=sqrt((0.51)^(2)+2^(2))~=2.06MeV` |
|
| 10. |
Sometimes a radioactive nucleus decays into a nucleus which itself is radioactive. An exmple is: `""^(38)"S" overset ("half-life") underset(=2.48) to ""^(38)Cl overset ("half-life") underset(=0.62h) to ""^(38)Ar ("stable")` Assume that we start with 1000 `""^(38)S` nuclei at time t=0. The number of `""^(30)Cl` is of count zero at t=0 and will again be zero at `t=oo`. At what value of t, would the number of counts be a maximum? |
|
Answer» The given decay sequence is `.^(38)("Sulper") overset ("half-life") underset(=2.48) to .^(38)Cl overset ("half-life") underset(=0.62h) to .(38)Ar ("stable")` At any time t, suppose `.^(38)S` has `N_(1) (t)` active nuclei and `.^(38)Cl` have `N_(2)(t)` active nuclei. `:. (dN_(1))/(dt)=-lambda_(1)N_(1)`=rate of formation of `.^(38)Cl` and `(dN_(2))/(dt)=-lambda_(2)N_(2)+lambda_(1)N_(1)`= net rate of decay of `.^(38)Cl=-lambda_(2)N_(2)+lambda_(1)N_(0)e^(-lambda_(1)t)` Multiplying both sides by `e^(lambda_(2)t)dt` and rearranging, `e^(lambda_(2)t)dN_(2)+lambda_(2)N_(2)e^(lambda_(2)t)d t=lambda_(1)N_(0)e^((lambda_(2)-lambda_(1)t)dt` Integrating both sides, we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^((lambda_(2)-lambda_(1)t)+C.......(i)` Where C is constant of integration. At `t=0, N_(2)=0 :. C=-(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))` Putting in (i), we get `N_(2)e^(lambda_(2)t)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^((lambda_(2)-lambda_(1)t))-1)]` `N_(2)=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1))[e^(-lambda_(1))-e^(-lambda_(2)t)]` for maximum count , `N_(2)=max, (dN_(2))/(dt)=0` `(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(1)t) (-lambda_(1))=(N_(0)lambda_(1))/(lambda_(2)-lambda_(1)) e^(-lambda_(2)t) (-lambda_(2))` or `(lambda_(1))/(lambda_(2))=(e^(-lambda_(2)t))/(e^(-lambda_(2)t))=e^((lambda_(1)-lambda_(2))t)` or `log_(e) ((lambda_(1))/(lambda_(2)))=(lambda_(1)-lambda_(2))tlog_(e) e` or `t=(log_(e)(lambda_(1)//lambda_(2)))/((lambda_(1)-lambda_(2)))=(log_(e) T_(2)//T_(1))/(0.693(1/(T_(1))-1/(T_(2))))` `t=(2.303log_(10) (0.62//2.48)xxT_(1)T_(2))/(0.693 (T_(2)-T_(1)))=(2.303(0-0.602)xx2.48xx0.62)/(0.693(0.62-2.48))=(2.303xx0.602xx2.48xx0.62)/(0.693xx1.86)=1.65sec` |
|
| 11. |
The isotope `_(5)^(12) B` having a mass `12.014 u`undergoes beta - decay to `_(6)^(12) C _(6)^(12) C `has an excited state of the nucleus `( _(6)^(12) C ^(**) at 4.041 MeV` above its ground state if `_(5)^(12)E` decay to `_(6)^(12) C ^(**) ` , the maximum kinetic energy of the `beta` - particle in unit of MeV is `(1 u = 931.5MeV//c^(2)` where c is the speed of light in vaccuum) .A. `9`B. `6`C. `3`D. `1` |
|
Answer» Correct Answer - A `._(5)B^(12)to._(6)C^(12)+._(-1)e^(0)+barv` Taking mass of `c^(12)` as 12.000u, mass defect `Deltam=12.014-12.000=0.014u` `Q=0.014xx931.5MeV =13.041 MeV` Out of this, 4.041MeV energy is taken by `._(6)C^(12**)` `:.` Max. K.E. of `beta` particle=13.041-4.041 `=9MeV` |
|
| 12. |
Deuteron is a bound state of a neutron and a proton with a binding energy `B=2.2MeV. A gamma`-ray of energy E is aimed at a deuteron nucleus to try to breack it into a (neutron +proton) such that the n and p move in the direction of the incident `gamma`-rays. If E=B, show that this can not happen. Hence, calculate how much bigger than B must E be for such a process to happen. |
|
Answer» Applying the principal of conservation of energy in the given situation, `E-B=K_(n)+K_(p)=(p_(n)^(2))/(2m)+(p_(p)^(2))/(2m).........(i)` for the principal of conservation of linear momentum, `p_(n)+p_(p)=E//c.......(ii)` If `E=B`, eqn. (i) would give `p_(n)=p_(p)=0`. Hence, the process cannot take place. for the process to take place, let `E=B+lambda`, where `lambda lt lt B`. form (i), `lambda=(p_(n)^(2))/(2m)+(p_(p)^(2))/(2m)` using (ii), `lambda=1/(2m) [p_(p)^(2)+(p_(p)-E//c)^(2)]` `:. 2p_(p)^(2)-(2E)/c p_(p)+((E^(2))/(c^(2))-2mlambda)=0 or p_(p)=(2E//c+-sqrt(4E^2//c^2-8((E^(2))/(c^(2))-2mlambda)))/4` for `p_(p)` to be real, the determinant on RHS must be positive. `:. (4E^2)/c^2-8((E^(2))/(c^(2))-2mlambda)ge0 or 16mlambda=(4E^2)/(c^2)` `lambda=(E^(2))/(4mc^(2))~=(B^(2))/(4mc^(2))` Hence, E must be bigger than B by `lambda=B^(2)//4mc^(2)`, for the given process to happen. |
|
| 13. |
A radioactive nucleus undergoes a series of decay according to the scheme `A overset(alpha)rarr A_(1) overset(beta^(-))rarr A_(2) overset(alpha)rarr A_(3)^(172) overset(gamma)rarr A_4`. |
| Answer» Emission of 2 alpha particles decreases the mass number by 8 and charge number 4. Emission of one `beta` particle increase the charge number by 1 without affecting mass number. `gamma` emission affects neither of two. Therefore, for `A_4`: mass no. =`180-8=172` charge no. =`72-4+1=69` | |
| 14. |
The sequence of decay of radioactive nucleus is `D overset alphato D_1oversetbeta toD_2 oversetalpha toD_3`. If nucleon number and atomic number of `D_2` are 176 and 71 respectively, what are their values for D and `D_3`? |
|
Answer» As `alpha` is `._2He^4` and `beta` is `._-1e^0`, therefore, for D, mass no. `=176+0+4=180` charge numeber `=71-1+2=72` for `D_3`, mass no. `= 176 -4=172` charge no. `=71-2=69` |
|
| 15. |
A radioavtive source in the form of a metal sphere of daimeter `10^(-3)` m emits `beta`-particles at a constant rate of `6.25 xx 10^(10)` particles per second. If the source is electrically insulated, how long will it take for its potential to rise by `1.0 V`, assuming that `80%` of the emitted `beta`-particles escape the socurce? |
|
Answer» As only `80%` of emitted `beta` particles escape form the source, therefore, Number of `beta` particles leaveing the source/sec `=80/100xx6.25xx10^(10)=5xx10^(10)` Positive charge gained/sec by the source `q=5xx10^(10)xx(1.6xx10^(-19))C=8xx10^(-9)C` Capacity of sphere, `C=4pi in_0r` `=1/(9xx10^9)((10^(-3))/2)=(10^(-12))/18F` Increase of potential of sphere per sec `=("Increase of charge"//"sec")/("Capacity")` `=8xx10^(-9)xx18/(10^(-12))=1.44xx10^5V` Time taken for rise of potential by one volt. `=1/(1.44xx10^5)=6.94xx10^(-6)s` |
|
| 16. |
The size of atomic nucleus is of the order of............ m and size of the atom is of the order of........... . |
| Answer» Correct Answer - `10^(-15); 10^(-10)m`, | |
| 17. |
Statement -1 : Large angle scattering of alpha particles led to the discovery of atomic nucleus. Statement -2 : Entire positive charge of atom is concentrated in the central core.A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
|
Answer» Correct Answer - A Both the statement are correct, and the latter is a correct explantion of the former. |
|
| 18. |
Assertion : `alpha`-particle is a helium nucleus. Reason : In `alpha`-decay, both the mass number as well as atomic number of the daugther is more than that of parent.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. When `alpha`-decay takes place, both the mass number and atomic number decreases. In general, `alpha` -decay is represented as `._(Z)X^(A)to._(Z-2)Y^(A-4)+._(2)He^(4)`+energy where X is the present nucleus and Y is the daugther nucleus. |
|
| 19. |
A nucleus `._nX^M` emits one `alpha` particle and one `beta`-particle. What are the mass number and atomic number of the product nucleus?A. `(m-4), n`B. `(m-4), (n-1)`C. `(m-3), (n+1)`D. `(m-3), (n-1)` |
| Answer» Correct Answer - B | |
| 20. |
How many electron potons and mass number in a nucleus of atomic number `11` and mass `24`? (i) number of electron = (ii)number of proton = (iii)number of neutrons = |
| Answer» In the nuclide `._11X^(24)`, number of protons `=11`, number of neutrons `=24-11=13`. Number of electrons=0, in the nucleaus. | |
| 21. |
Why is density of a nucleus much more than the atomic density? |
| Answer» This is because nuclear size is very small `(~=10^(-14)m)` and atomic size is very large `(~=10^(-10)m)`. And atomic mass is only slightly greater than the nuclear mass, due to additional mass of electrons. | |
| 22. |
If one a.m.u.=`1.66xx10^(-27)kg`, when is the mass of one atom of `C^(12)`? |
|
Answer» As `1 a.m.u. =1/12` mass of one atom of `C^(12)` `:.` mass of one atom of `C^(12)=12 a.m.u` `=12xx1.66xx10^(-27)kg=1.992xx10^(-26)kg.` |
|
| 23. |
The binding energy per nucleon for `._(6)C^(12) is 7.68 MeV//N` and that for `._(6)C^(13) is 7.47MeV//N`. Calculate the energy required to remove a neutron form `._(6)C^(13)` |
|
Answer» Correct Answer - 4.95 MeV Total B.E. of `._(6)C^(12)=12xx7.68=92.16 MeV` Total B.E. of `._(6)C^(13)=13xx7.47=97.11 MeV` `._(6)C^(13)` has one excess neutron than `._(6)C^(12)`, `:.` Energy required to remove a neutron `=97.11-92.16=4.95MeV` |
|
| 24. |
Consider the fission `._(92)U^(238)` by fast neutrons. In one fission event, no neutrons are emitted and the final stable and products, after the beta decay of the primary fragments are `._(58)Ce^(140)` and `._(44)Ru^(99)`. Calculate Q for this fission process, The relevant atomic and particle masses are: `m(._(92)U^(238))=238.05079u, m(._(58)Ce^(140))=139.90543 u, m(._(34)Ru^(99))=98.90594 u` |
|
Answer» for the fission reaction, `._(92)U^(238)+._0n^1to._(58)Ce^(140)+._(34)Ru^(99)+Q` mass defect, `Deltam= "mass of" U^(238)+"mass of n-(mass of" Ce^(140)+"mass" Ru^(99))` `=238.05079+1.00867-(139.90543+98.90594)=0.24809u` `Q=0.24809xx931MeV=230.97MeV` |
|
| 25. |
A sample of a radioactive element has a mass of `10 g` at an instant `t=0`. The approxiamte mass of this element in the sample after two mean lives is .A. `1.35g`B. `2.50g`C. `3.70g`D. `6.30g` |
|
Answer» Correct Answer - A `t=2tau=2xx1.44T=2.88T` As `N/(N_(0))=m/(m_(0)) =(1/2)^(t//T)=(1/2)^(2.38)` `:. m=m_(0)(1/2)^(2.88)=10(1/2)^(2.88)g` `=10xx0.135g=1.35g` |
|
| 26. |
The half life of a radioactive substance is `20` minutes . The approximate time interval `(t_(2) - t_(1))` between the time `t_(2)` when `(2)/(3)` of it had decayed and time `t_(1)` when `(1)/(3)` of it had decay isA. `14 mi n`B. `20 mi n`C. `28 mi n`D. `7mi n` |
|
Answer» Correct Answer - B Here, `T=20 "minutes"` In time `t_(1)`, the radioactive sample left behind `=1-1/3=2/3`. In time `t_(2)`, the radioactive sample left behind `=1-1/3=2/3`. As `n/(N_(0))=(1/2)^(t//T) :. 2/3(1/2)^(t_(1)//T).........(i)` and `1/3=(1/2)^(t_(2)//T)` Dividing (ii) by (i), we get `1/2=(1/2)^((t_(2)-t_(1))//T)` or `t_(2)-t_(1)=T=20min`. |
|
| 27. |
A radio isotope `X` with a half-life `1.4 xx 10^9` years decays of `Y` which is stable. A sample of the rock from a cave was found to contain `X` and `Y` in the ratio `1 : 7`. The age of the rock is.A. `1.96xx10^(9)yr`B. `3.92xx10^(9)yr`C. `4.20xx10^(9)yr`D. `8.40xx10^(9)yr` |
|
Answer» Correct Answer - C Here, `X/Y=1/7 :. X/(X+Y)=1/(1+7)=1/8=1/(2^(3))=(1/2)^(3)` `n=3 or t/T=3` `t=3T=3xx1.4xx10^(9)yrs =4.20xx10^(9)yrs` |
|
| 28. |
Two samples X and Y contain equal amounts of radioactive substances. If `1/16 th` of a sample X and `1/256 th` of sample Y remain after 8h, then the ratio of half periods of X and Y isA. `2:1`B. `1:2`C. `1:4`D. `1:16` |
|
Answer» Correct Answer - A Here, `t=8h, T_(1)//T_(2)=?` `(A_(1))/(A_(0))=(N_(1))/(N_(0))=(1/2)^(n_(1))=1/16=(1/2)^(4) :. n_(1)=4` `(A_(2))/(A_(0))=(N_(2))/(N_(0))=(1/2)^(n_(2))=1/256=(1/2)^(8) :. n_(2)=8` `(T_(1))/(T_(2)) =(t//n_(1))/(t//n_(2))=(n_(2))/(n_(1))=8/4=2:1` |
|
| 29. |
The decay constant of radio isotope is `lambda`. If `A_(1) and A_(2)` are its activities at times `t_(1) and t_(2)` respectively, the number of nuclei which have decayed during the time `(t_(1)-t_(2))`A. `A_(1)t_(1)-A_(2)t_(2)`B. `A_(1)-A_(2)`C. `(A_(1)-A_(2))//lambda`D. `lambda(A_(1)-A_(2))` |
|
Answer» Correct Answer - C Activity at time `t_(1), A_(1)=lambdaN_(1) or N_(1)A_(1)//lambda` Activity at time `t_(2), A_(2)=lambdaN_(2) or N_(2)A_(2)//lambda` Therefore, the number of nuclei decayed during time interval `(t_(1)-t_(2))` is `N_(1)-N_(2)=(A_(1))/lambda-(A_(2))/lambda=((A_(1)-A_(2))/lambda)` |
|
| 30. |
A radioactive isotope has a life of T years. How long will it take the activity to reduce to (a) `1%` of its original activity?A. `3.2T yr`B. `4.6T yr`C. `6.6T yr`D. `9.2T yr` |
|
Answer» Correct Answer - C form `A=A_(0)=N/(N_(0)) =(1/2)^(n) =1/100` `:. n(log1-log2)=log1-log100` `n(0-0.3010)=0-2` `n=2/0.3010=6.6` `t=nT=6.6Tyrs` |
|
| 31. |
Calculate and compare the energy released by (a) fusion of 1.0kg of hydrogen deep within the sun, and (b) the fission of 1.0kg of `U^(235)` in a fission reactor. |
|
Answer» In sun, four hydrogen nuclie fuse to form a helium nucleus with the release of 26MeV energy. `:.` Energy released by fusion of 1kg of hydrogen `=(6xx10^(23)xx10^3)/4xx26MeV` `E_1=39xx10^(26)MeV` As energy released in fission of one atom of `._(92)U^(235)=200MeV`, `:.` Energy released by fusion of 1kg of `._(92)U^(235)=(6xx10^(23)xx10^3)/235xx200 MeV` `E_2=5.1xx10^(26)MeV` `(E_1)/(E_2)=(39xx10^(26))/(5.1xx10^(26))=7.65`, i.e. energy released in fusion is 7.65 times the energy released in fission. |
|
| 32. |
This question contains Statement - 1 and Statement -2 Of the four choice given after the Statements , choose the one that best decribes the two Statements Statement- 1: Energy is released when heavy undergo fission or light nuclei undergo fusion and Statement- 2: for nuclei , Binding energy nucleon increases with increasing `Z` while for light nuclei it decreases with increasing `Z`A. Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, but Statement-2 is falseD. Statement-1 is false, but Statement-2 is true. |
|
Answer» Correct Answer - A Both the statements 1 and 2 are true, and statement 2 is correct explanation of statement1. |
|
| 33. |
A hypothetica atom has energy levels uniformaly separated by 1.3 eV. At a temperature 2500k, what is the ratio of number of atoms in 15th excited state to the number in 13 th excited state. |
|
Answer» It is known that in excited state x, no. of atoms, `N_x=N_0e^((E_x-E_0)//kT)` `:. (N_(15))/(N_(13))=e^(-(E_(15)-E_(13))//kT)` `=exp.[-(2xx1.3eV)/(kT)]` `=exp.[-(2.6xx1.6xx10^(-19))/(1.38xx10^(-23)xx2500)]` `(N_(15))/(N_(13))=exp.[-12]=1/(e^(12))=6.15xx10^-6` |
|
| 34. |
Fig. shows energy level diagram of hydrogen atom. Find out the transition which results in the emission of a photon of wavelength 496 nm. Which transition corresponds to emission of radiation of maximum wavelength? Justify your answer. |
|
Answer» Correct Answer - `E_(4)toE_(2) : 4 to 3` Energy levels of hydrogen atom are `E_(1)=-13.6eV, E_(2)=-3.4eV` `E_(3)=-1.51eV, E_(4)=-0.85eV` Energy of a photon of wavelength 496nm is `E=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(496xx10^(-9)(1.6xx10^(-19)))eV=2.5eV` As `E_(4)-E_(2)=-0.85+3.4=2.55eV=E`, therefore, the transition form n=4 to n=2 level results in the emission of photon of wavelength 496 nm. Further, wavelength emitted will be maximum, when energy emitted is minimum. The transition n=4 to n=3 level will give maximum wavelength as energy emitted is minimum. |
|
| 35. |
Shows the energy levels for an electron in a certain atom. Which transition shown represented the emission of a photon with the most energy? A. IIB. IC. IVD. III |
|
Answer» Correct Answer - D The photon emitted is with maximum energy in case of transition III. `E=E_(2)-E_(1)=(-13.6)/(2^(2))-(-13.6)/(1^(2))=10.2eV` |
|
| 36. |
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits form the upper level to the lower level. |
|
Answer» Here, `E=2.3 eV =2.3xx 1.6xx10^(-19)J.` As `E=hv= :. v=E/h=(2.3xx1.6xx10^(-19))/(6.6xx10^(-34))=5.6xx10^(14)Hz` |
|
| 37. |
The energy levels of an atom are shown in fig. Which transition corresponds to emission of radiation of (i) maximum wavelength (ii) minimum wavelength? |
|
Answer» As energy emitted, `E=(hc)/lambda prop1/lambda` ` :.` for maximum wavelength , E should be minimum. Transition A for which energy diff. E is minimum corresponds to emission of radiation of max. wavelength. Similarly, transition D for which energy diff. E is maximum, corresponds to emission of radiation of minimum wavelength. |
|
| 38. |
Find the wavelength of `H_(alpha)` line given the value of Rydberg constant, `R=1.1xx10^7m^-1`. |
|
Answer» `H_(alpha)` line is first line of Balmer series for which n=3 `lambda=R(1/(2^2)-1/(3^2))=Rxx5/36` `lambda=36/(5R)=36/(5xx1.1xx10^7)m` `=6.545xx10^-7m` |
|
| 39. |
The radiation corresponding to ` 3 rarr 2` transition of hydrogen atom falls on a metal surface to produce photoelectrons . These electrons are made to enter circuit a magnetic field `3 xx 10^(-4) T` if the ratio of the largest circular path follow by these electron is `10.0 mm , the work function of the metal is close toA. `0.8eV`B. `2.14eV`C. `1.8eV`D. `1.1eV` |
|
Answer» Correct Answer - D `1/2mv^(2)=eV or mv=sqrt(2meV)` Radius of circular path of electron `r=(mv)/(qB)=(sqrt(2meV))/(eB) =1/B sqrt((2mV)/e)` or `V=(B^(2)er^(2))/(2m)` `=((3xx10^(-4))^(2)xx(1.6xx10^(-19))xx(10xx10^(-3))^(2))/(2xx(9xx10^(-31)))` `=0.8V` KE of electron, K=eV=0.8eV for transition between 3 to 2, we have `E=13.6(1/(2^(2))-1/(3^(2)))=(13.6xx5)/36=1.88eV` Work fuction, `phi_(0)=E-K=1.88-0.8` `=1.08eV~~1.1eV` |
|
| 40. |
Calculate the equivalent energy of electrons and proton at rest. Given that mass of electron `=9.1xx10^(-31)kg` and mass of proton `=1.673xx10^(-27)kg`. |
|
Answer» Correct Answer - 0.511 MeV, 941.1 MeV `E_(1)=m_(e)c^(2)=9.1xx10^(-31)(3xx10^(8))^(2)J` `=(1.673xx9xx10^(-11))/(1.6xx10^(-13)) MeV=941.1MeV` |
|
| 41. |
The mass of a H-atm is less than the sun f the masses of a proton and electron. Why is this? |
|
Answer» According to mass energy equivalence established by Einstein, `E = mc^(2)`. If B represents binding energy of hydrogen atom `(=13.6 eV)`, the equivalent mass of the energy `= B//c^(2)`. Hence, mass of a H-atom `= m_(p) + m_(e) - B//c^(2)`. It is less than sum of the masses of a porton and an electron. |
|
| 42. |
On bombardment of `U^235` by slow neutrons, `200 MeV` energy is released. If the power output of atomic reactor is `1.6 MW`, then the rate of fission will beA. `5xx10^(10)`B. `5xx10^(12)`C. `5xx10^(4)`D. `5xx10^(16)` |
|
Answer» Correct Answer - D Output power `1.6MW=1.6xx10^(6)J//s` Energy released/fission =200MeV `=200xx1.6xx10^(-13)J` `:.` Nuclear of fission/sec `=(1.6xx10^(6))/(200xx1.6xx19^(-13))=5xx10^(16)` |
|
| 43. |
Electron capture occurs more often than positron emission in heavy elements. Heavy elements exhibit radioactivity.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 - B Electron capture occurs more often than positron emission in heavy elements. This is because if positron emission is energetically allowed, electron capture is necessarily allowed, but the reverse is not true i.e., when electron capture is energetically allowed, positron emission is not necessarily allowed. So, the assertion is true, but reason is not a correct explanation of the assertion. |
|
| 44. |
Imagine removing one electron form `He^(4)` and `He^(3)`. Their energy levels, as worked out on the basis of bohr model will be very close. Explain why. |
| Answer» Imagine one electorn removed form `._2He^(4) and _2He^(3)`. As both the residual nuclei are very heavy compared to the mass of electron removed, therefore, their energy levels, worked out on the basis of bohr model, will be very close. | |
| 45. |
If a star can convert all the He nuclei completely into oxygen nuclei. The energy released per oxygen nuclei is (Mass of the helium nucleus is `4.0026` amu and mass of oxygen nucleus is `15.9994` amu)A. `7.6MeV`B. `56.12MeV`C. `10.24MeV`D. `23.9MeV` |
|
Answer» Correct Answer - C When four helium nuclei are fused together, one oxygen nucleus is formed. The reaction is `4xx2He^(4)to._(8)O^(16)+Q` Mass defect, `Deltam=4xxm_(He)-m_(O_(2))` `=4xx4.0026-15.9994=0.011a.m.u.` Energy released `E=0.011xx931=10.24MeV` |
|
| 46. |
One Mev positron encounters one MeV electron travelling in opposite direction. What is the wavelength of photons produced, given rest mass energy of electron or positron `=0.512MeV` ? Take `h=6.62xx10^(-34)J-s`. |
|
Answer» Correct Answer - `8.2xx10^(-13)m` The nuclear reaction is `._(+1)e^(0)+._(-1)e^(0)to 2gamma` Total energy before reaction =`1.512xx2MeV` After reaction, energy of each photon, `E=(1.512xx2)/2MeV` `E=1.512xx1.6xx10^(-13)J=(hc)/lambda` `lambda=(hc)/(1.512xx1.6xx10^(-13))=(6.6xx10^(-34)xx3xx10^(8))/(1.512xx1.6xx10^(-13))` `=8.2xx10^(-13)m` |
|
| 47. |
A set of atoms in an excited state decays.A. in general to any of the states with lower energyB. into a lower state only when excited by an external electric fieldC. all together simultaneously into a lower stateD. to emit photons only when they collide |
|
Answer» Correct Answer - A A set of atoms in an excited state decays in general to any of the states with lower energy. Choice (a) Is correct. |
|
| 48. |
The Bohr model for the spectra of a H-atomA. will not be applicable to hydrogen in the molecular formB. will not be applicable as it is for a H-atomC. is valid only at room temperatureD. predicts continuous as well as discrete spectral lines |
|
Answer» Correct Answer - A::B Bohr model for the spectra of H-atom is not applicable to hydrogen in the molecular form. As its is, the Bohr model is not applicable to a He atom. These are the known limitations of Bohr atom model. Choices (a) and (b) are correct. |
|
| 49. |
Assuming that four hydrogen atom combine to form a helium atom and two positrons, each of mass 0.00549u, calculate the energy released. Given `m(._(1)H^(1))=1.007825u, m(._(2)He^(4))=4.002604u`. |
|
Answer» Correct Answer - `25.7MeV` The nuclear reaction is `4._(1)H^(1) to ._(2)He^(4)+2 ._(+1)e^(0)+Q` Total initial mass `=4xx1.007825=4.031300u` Total final mass `=4.002604+2xx0.000549` `=4.003702u` Mass defect, `Deltam=4.031300-4.003702` `0.027598u` Total energy released `=0.027598xx931MeV` `=25.7MeV` |
|
| 50. |
Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton can not combine to produce a H-atom,A. because of energy conservationB. without simultaneously releasing energy in the form of radiationC. because of momentum conservationD. because of angular momentum conservation |
|
Answer» Correct Answer - A::B When a beam of free electrons is aimed towards free protons, the electrons get scattered on account of energy conservation. An electron and a proton can cambine to produce a H-atom only if they simultaneously release energy in the form of radiation. Choices (a) and (b) are correct. |
|