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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. |
During the transformation of `._(c )X^(a)` to `._(d)Y^(b)` the number of `beta`-particles emitted are a. `d + ((a - b)/(2)) - c` b. `(a - b)/(c )` c. `d + ((a - b)/(2)) + c` d. `2c - d + a = b`A. a. `d + ((a - b)/(2)) - c`B. b. `(a - b)/(c )`C. c. `d + ((a - b)/(2)) + c`D. d. `2c - d + a = b` |
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Answer» Correct Answer - a. `d + ((a - b)/(2)) - c` a. For transformation of `._(c )X^(a) rarr ._(d)Y^(b) + x_(2)He^(4) + beta` Number of `alpha`-particles emitted `= (a - b)/(4)` Equating for atomic number on both sides, we get `c = d + 2 ((a - b)/(2)) + beta (-1)` or `beta = d + (a - b)/(2) - c` |
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| 2. |
During the transformation of `.^(b)X_(a) rarr .^(d)Y_(c)` the number of `beta`-particles emitted isA. `((b -d))/(4)`B. `(c -a) + (1)/(2) (b -d)`C. `(a -c) - (1)/(2) ( b-d)`D. `(b -d) + 2 (c -a)` |
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Answer» Correct Answer - B `._(a)^(b) X rarr ._(c)^(d)Y + x_(1)^(4) alpha + y_(-1)^(0)beta` `b = d + 4x + 0y " " a = c + 2x - y` or, `b =d + 4x " " or, a = c + 2.(b-d)/(4) -y` or, `x = (b-d)/(4) " " or, a = c + (b -d)/(2) - y` or, `y = (c -a) + ((b -d))/(2)` |
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| 3. |
How many `alpha`-particles are emitted in the nuclear transformation: `._(84)Po^(215) rarr ._(82)Pb^(211) + ? ._(2)He^(4)` |
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Answer» Correct Answer - B Let `x alpha`-particles are emitted. `._(84)Po^(215) rarr ._(82)Pb^(211) + x ._(2)He^(4)` Equating atomic mass of the both sides `215 = 211 + 4x` `:. X = (215 - 211)/(4) = (4)/(4) = 1` Therefore, one `alpha`-particle is emitted. |
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| 4. |
Which one of the following nuclear transformation is `(np)` type? a. `._(3)Li^(7) + ._(1)H^(1) rarr ._(4)Be^(7) + ._(0)n^(1)` b. `._(33)As^(75) + ._(5)He^(4) rarr ._(35)Bi^(78) + ._(0)n^(1)` c. `._(83)Bi^(209) + ._(1)H^(2) rarr ._(84)Po^(210) + ._(0)n^(1)` d. `._(21)Sc^(45) + ._(0)n^(1) rarr ._(20)Ca^(45) + ._(1)H^(1)`A. a. `._(3)Li^(7) + ._(1)H^(1) rarr ._(4)Be^(7) + ._(0)n^(1)`B. b. `._(33)As^(75) + ._(5)He^(4) rarr ._(35)Bi^(78) + ._(0)n^(1)`C. c. `._(83)Bi^(209) + ._(1)H^(2) rarr ._(84)Po^(210) + ._(0)n^(1)`D. d. `._(21)Sc^(45) + ._(0)n^(1) rarr ._(20)Ca^(45) + ._(1)H^(1)` |
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Answer» Correct Answer - d. `._(21)Sc^(45) + ._(0)n^(1) rarr ._(20)Ca^(45) + ._(1)H^(1)` In reaction `._(21)Sc^(45) + ._(0)n^(1) rarr ._(20)Ca^(45) + ._(1)H^(1)` Neutron is attacking while proton is produced along with the product. |
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| 5. |
After np orbitals are filled, the next orbital filled will be :-(a). `(n+1)s`(b). `(n+2)p`(c). `(n+1)d`(d). `(n+2)s`A. `(n+1) s`B. `(n+2)p`C. `(n+1) d`D. `(n+2) s` |
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Answer» Correct Answer - A Total number of electrons in an orbital `=2(2l+1)` The value of `l` varies from `0` to `n-1` :. Total numbers of electrons in any orbit `=sum_(l=0)^(l=n-1) 2(2l+1)`. |
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| 6. |
The total number of `alpha`- and `beta`-particles given out during nuclear given nuclear transformation is:A. `2`B. `4`C. `6`D. `8` |
| Answer» Correct Answer - D | |
| 7. |
Which one of the following nuclear transformation is `(np)` type? a. `._(3)Li^(7) + ._(1)H^(1) rarr ._(4)Be^(7) + ._(0)n^(1)` b. `._(33)As^(75) + ._(5)He^(4) rarr ._(35)Bi^(78) + ._(0)n^(1)` c. `._(83)Bi^(209) + ._(1)H^(2) rarr ._(84)Po^(210) + ._(0)n^(1)` d. `._(21)Sc^(45) + ._(0)n^(1) rarr ._(20)Ca^(45) + ._(1)H^(1)`A. `._(3)Li^(7) + ._(1)H^(1) rarr ._(4)Be^(7) + ._(0)n^(1)`B. `._(33)As^(75) + ._(2)He^(2) rarr ._(35)Br^(78) + ._(0)n^(1)`C. `._(83)Bi^(209) + ._(1)H^(2) rarr ._(84)Po^(210) + ._(0)n^(1)`D. `._(21)Sc^(45) + ._(0)n^(1) rarr ._(20)Ca^(45) + ._(1)H^(1)` |
| Answer» Correct Answer - D | |
| 8. |
The reaction `._(1)D^(2) + ._(1)T^(3) rarr ._(1)He^(2) + ._(0)n^(1)` is an example ofA. Nuclear fissionB. Nuclear fusionC. Artifical radioactivityD. Radioactive disintegration |
| Answer» Correct Answer - B | |
| 9. |
The equipment used to carry out nuclear reaction in a controlled manner is calledA. Breeder reactorB. Nuclear reactorC. Thermonuclear fissionD. Cyclotron |
| Answer» Correct Answer - B | |
| 10. |
The number of `alpha and beta-`particles emitted in the nuclear reaction `._(90)Th^(228) rarr ._(83)Bi^(212)` are respectivelyA. 4 , 1B. 3 , 7C. 8, 1D. 4, 7 |
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Answer» Correct Answer - A `._(90)Th^(228)rarr ._(83)Bi^(212)` No. of `alpha-"particle" = (228 - 212)/(4) = (16)//(4) = 4` No. of `beta-"particle" = 2xx 4 - 90 + 83` |
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| 11. |
The number of `alpha-` particles emitted per second by `1g` of `Ra^(226)` is `3.7 xx 10^(10)`. The decay constant isA. `1.39xx10^(-11)s^(-1)`B. `13.9 xx10^(-11)s^(-1)`C. `139xx10^(-11)s^(-1)`D. `0.139xx10^(-11)s^(-1)` |
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Answer» Correct Answer - A `(Number of at oms d i s i ntegrati ng per se c o nd )/(Total n umber of at oms present)=K` or `(3.7xx10^(10))/((6.02xx10^(23))/(226))=(226xx3.7xx10^(10))/(6.02xx10^(23))=K` |
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| 12. |
Radioactivity of a radioactive element remains `(1)/(10)` of the original radioactivity after 2.303 seconds. The half-life period isA. 2.303B. 0.2303C. 0.693D. 0.0693 |
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Answer» Correct Answer - C `2.303 = (2.303)/(0.693) xx t_(1//2) log 10` |
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| 13. |
The half-life period of Uranium is 4.5 billion years. After 90 billion years, the number of moles of Helium liberated from the following nuclear reaction will be `._(92)U^(238) rarr ._(90)Th^(234) + ._(2)He^(4)`A. 0.75 moleB. 1.0 moleC. 11.2 moleD. 22.4 mole |
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Answer» Correct Answer - A `N = (N_(0))/(2^(n)) = N = (N_(0))/(2^(2)) = (1)/(4)` `:.` converted moles `= 1 - (1)/(4) = (3)/(4) = 0.75` |
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| 14. |
Radioactivity of a radioactive element remains `1//10` of the original radioactivity after `2.303` seconds. The half life period isA. `2.303`B. `0.2303`C. `693`D. `0.693` |
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Answer» Correct Answer - D `K=(2.303)/(t)log``(a)/(a-x)` `[{:(t=2.303(given)),((a-x)=(1)/(10)a_(0)):}]` `=(2.303)/(t)log``(1)/(a//10)=1` `:. t_(1//2)=(0.693)/(K)=(0.693)/(1)=0.693` |
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| 15. |
The half-life period of `._(53)I^(125)` is 60 days. What percent of radioactivity would be present after 180 daysA. 0.25B. `12.5%`C. 0.5D. 0.75 |
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Answer» Correct Answer - B `N = (N_(0))/(2^(n)) n = (180)/(60) = 3` `(N)/(N_(0)) = (1)/(2^(3)) rArr (N)/(N_(0)) xx 100 = (1)/(8) xx 100 = 12.5%` |
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| 16. |
Who observed that when the nucleus of uranium atom was bombarded with fast moving neutrons, it becomes so unstable that it is immediately broken into two nuclei of nearly equal mass besides other fragmentsA. J.J. ThomsonB. ChadwickC. EinsteinD. Hahn and Strassmann |
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Answer» Correct Answer - D Hahn and Strassmann discovered the phenomenon of nuclear fission in 1939 |
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| 17. |
The nuclei of two radioactive isotopes of same substance `A^(236)` and `A^(234)` are present in the ratio of `4:1` in an ore obtained from some other planet. Their half lives are `30min` and `60min` respectively. Both isotopes are alpha emitters and the activity of the isotope with half-life `30min` is `10^(6)dps`. Calculate after how much time their activites will become identical. Also calculate the time required to bring the ratio of their atoms to `1:1`.A. 30 minB. 60 minC. 120 minD. 150 min |
| Answer» Correct Answer - c | |
| 18. |
The age of most ancient geological formations is estimated bya)`C-14` dating methodb)`K-Ag` methodc)`U-Pb` methodd)`Ra-Rn` methodA. potassium argon methodB. carbon 14 dating method `C. radium silicon methodD. uranium lead method |
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Answer» Correct Answer - D The age of most ancient geological objects is determined by uranium lead method. This methos is based on the radoactive porperties of uranium. In theis methos,we assume that initially the rock did not contain any lead , we can calculate the age of the rock by measuring the ratio of concentrations of `_(92)U^(238)nd _(82)Pb^(206)` |
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| 19. |
If velocity of an electron in `I` orbit of `H` atom is `V`, what will be the velocity of electron in `3^(rd)` orbit of `Li^(2+)`A. `V`B. `V//3`C. `3V`D. `9V` |
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Answer» Correct Answer - A `V=2.188xx10^(6) Z/n m//s` Now, `V prop Z/n` so, `V_(Li^(2+))/V_(H)=-(Z_(1)//n_(1))/(Z_(2)//n_(2))=(3//3)/(1//1)=1` or, `V_(Li^(2+))=V_(H)` |
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| 20. |
Which is the correct relationship?(a). `E_(1)` of `H=1//2E_(2)` of `He^(+)=1//3E_(3)`of `Li^(2+)=1//4E_(4)`of `Be^(3+)`(b). `E_(1)(H)=E_(2)(He^(+))=E_(3)(Li^(2+))=E_(4)(Be^(3+))`(c). `E_(1)(H)=2E_(2)(He^(+))=3E_(3)(Li^(2+))=4E_(4)(Be^(3+))`(d). No relationA. `E_(1) "of" H=1//2 E_(2) "of" He^(+)=1//3 E_(3) "of" Li^(2+)1//4 E_(4) "of" Be^(3+)`B. `E_(1)(H)=E_(2)(He^(+))=E_(3)(Li^(2+))=E_(4)(Be^(3+))`C. `E_(1)(H)=2E_(2)(He^(+))=3E_(3)(Li^(2+))=4E_(4)(Be^(3+))`D. No relation |
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Answer» Correct Answer - B `E_(1)(H)=-13.6xx1^(2)/1^(2)=-13.6 eV" " , " "E_(2)(He^(+))=-13.6xx2^(2)/2^(2)=-13.6 eV` `E_(3)(Li^(2+))=-13.6xx3^(2)/3^(2)=-13.6 eV " " , " " E_(4)(Be^(3+))=-13.6xx4^(2)/4^(2)=-13.6 eV` `:. E_(1)(H)=E_(2)(He^(+))=E_(3)(Li^(2+))=E_(4)(Be^(3+))` |
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| 21. |
(a) On analysis a sample of uranium ore was found to contain `0.277g` of `._(82)Pb^(206)` and `1.667g` of `._(92)U^(238)`. The half life period of `U^(238)` is `4.51xx10^(9)` year. If all the lead were assumed to have come from decay of `._(92)U^(238)`, What is the age of earth? (b) An ore of `._(92)U^(238)` is found to contain `._(92)U^(238)` and `._(82)Pb^(206)` in the weight ratio of `1: 0.1` The half-life period of `._(92)U^(238)` is `4.5xx10^(9)` year. Calculate the age of ore. |
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Answer» (a) given `""""_(92)""^(238)U= 1.667 g = (1.667)/(238) " mole" = 7.004 xx 10""^(-3) " mole" """"_(82)""^(206)pb = 0.227 g = (0.277)/(206) = 1.345 xx 10""^(-3) " mole"` In this case all the lead comes from the decay of uranium `therefore`No. of mole of Pb formed `= (0.277)/(206) = 1.345 xx 10""^(-3) " mole"` No. of mole of U decayed= `1.345 xx 10""^(-3) " mole"` Initial no of mole of U before decay `= 7.004 xx 10""^(-3) + 1.345 xx 10""^(-3)` Moles of U after decay `N = 7.004 xx 10""^(-3) " year"""^(-1)` `lambda = (0.693)/(t""_(0.5)) = (0.693)/(4.5 xx 10""^(9)) = 1.536 xx 10""^(-10) " year"""^(-1)` As we know `t = (2.303)/(lambda) "log" (N""_(0))/(N) = (2.303)/(1.536 xx 10""^(10)) "log" (8.349 xx 10""^(-3))/(7.004 xx 10""^(-3))` `therefore` Age of earth `(t) = 46.85 xx 10""^(-4) " mole"` b) `N""_(0)=(1)/(238)+(0.1)/(206)=46.85xx10""^(-4)" mole"` `N=(1)/(238)=42.0xx10""^(-4)" mole"` `lamda=(0.693)/(4.5xx10""^(9))=1.54xx10""^(-4)year""^(-1)` `t=(2.303)/(1.54xx10""^(10))log.(46.85xx10""^(-4))/(42.0xx10""^(-4))` `t=7.097xx10""^(8)year` Note: Let a radioactive element A decay to daughter element B A to B `lamda""_(A)" and "lamda""_(B)" are decay constant of A and B `" `"Maximum activity time of daughter element can be"` `"calculated as "t""_(max)=(2.303)/((lamda""_(B)-lamda""_(A)))log""_(10)[lamda""_(B)/lamda""_(A)]` |
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| 22. |
S : When n/p ratio is high, nuclei emits `beta-` particles E : The nuclear change is : `nto p""^(+)+e""^(-)+overset(-)v`A. Both S and E are correct and E is cporrect explaination of SB. Both S and E are correct and E not correct explaination of SC. S is correct but E is wrongD. S is wrong but the E is correct |
| Answer» Correct Answer - a | |
| 23. |
Which emits `beta-`particlesA. `._(1)H^(3)`B. `._(6)C^(14)`C. `._(19)K^(40)`D. All |
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Answer» Correct Answer - D Because all are radioactive isotopes and can emit `alpha , beta " &" gamma`-particles. |
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| 24. |
Consider the following nuclear reaction, `._(92)^(238)M rarr ._(y)^(x)N + 2 ._(2)^(4)He ._(y)^(x) N rarr ._(B)^(A) L + 2beta^(+)`A. 140B. 144C. 142D. 146 |
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Answer» Correct Answer - B `._(92)M^(238) rarr ._(y)N^(x) + 2 ._(2)He^(4)` `._(y)N^(x) rarr ._(B)L^(A) + 2 beta^(+)` `._(y)N^(230) overset((92 - 2 xx2)rarr N^((238 - 4 xx 2)) = ._(88)N^(230)` `._(88)N^(230) overset(2beta^(+))rarr ._((88 -2))L^((230)) = ._(86)L^(230)` Total no of neutrons in `._(86)L^(230)` `230 - 86 = 144` |
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| 25. |
Consider the following nuclear reactions `._(92)^(238)M rarr ._(X)^(Y)N + 2alpha, ._(X)^(Y)N rarr ._(B)^(A)L + 2beta` The number of neutrons in the element L isA. 142B. 144C. 140D. 146 |
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Answer» Correct Answer - C Nuclear reaction `._(92)^(238)M rarr ._(X)^(Y)N + 2 alpha (2 xx ._(2)^(4)He) or ._(88)^(230)N` `._(88)^(230)N rarr ._(B)^(A)L + 2 ._(+1)^(0)e or ._(90)^(23)L` `.:` No of neutron in L `= 230 - 86 = 140` |
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| 26. |
Calculate the energy released in joules and `MeV` in the following nuclear reaction: `._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1)` Assume that the masses of `._(1)H^(2)`, `._(2)He^(3)`, and neutron `(n)`, respectively, are 2.40, 3.0160, and 1.0087 in amu. |
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Answer» `Delta m [2 xx 2.0141] - 3.0160 - 1.0087 = 3.5 xx 10^(-3) "amu"` `:. Delta E = Delta m xx 931.478 MeV` `= 3.5 xx 10^(-3) xx 931.478 = 3.260 MeV` `:. Delta E = 5.223 xx 10^(-13) J` |
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| 27. |
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [`a_0` is Bohr radius] :A. `h^(2)/(4pi^(2)ma_(0)^(2))`B. `h^(2)/(16 pi^(2)ma_(0)^(2))`C. `h^(2)/(32pi^(2)ma_(0)^(2))`D. `h^(2)/(64pi^(2) ma_(0)^(2))` |
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Answer» Correct Answer - C `mv (4a_(0))=h/pi` so, `v=h/(4mpia_(0))` so `KE=1/2 mv^(2)=1/2 m. (h^(2))/(16m^(2)pi^(2)a_(0)^(2))=(h^(2))/(32mpi^(2) a_(0)^(2))` |
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| 28. |
An electron in an atom jumps in such a way that its kinetic energy changes from x to `x/4`. The change in potential energy will be:A. `+3/2 x`B. `-3/8 x`C. `+3/4 x`D. `-3/4 x` |
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Answer» Correct Answer - 1 Change in `P.E. =-(2x)/(4)+(2x)rArr 3/2 x` |
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| 29. |
For an electron, with n=3 has only one radial node. The orbital angular momentum of the electron will be :A. `0`B. `sqrt(6) h/(2pi)`C. `sqrt(2) h/(2pi)`D. `3(h/(2pi))` |
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Answer» Correct Answer - C Number of radial nodes `=nl-1=1, n=3 :. l=1`. Orbital angular momentum `=sqrt(l(l+1)) (h)/(2pi)=sqrt(2) (h)/(2pi)` |
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| 30. |
The change in orbital angular momentum corresponding to an electron transition inside a hydrogen atom can be-(a). `(h)/(4pi)`(b). `(h)/(pi)`(c). `(h)/(2pi)`(d). `(h)/(8pi)`A. `h/(4pi)`B. `h/pi`C. `h/(2pi)`D. `h/(8pi)` |
| Answer» Correct Answer - B, C | |
| 31. |
Electrons in the H-atoms jumps from some higher level to 3rd energy level . If six spectral lines are possible for the transition find the initial position of electron. |
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Answer» Correct Answer - 6 `x+e^(-) rarr x^(-)` energy released =`E.A_(1)+E.A_(2)=30.87 eV//atom` Let no. of moles of `X` be a `:. axxN_(a)xx30.87=6xxN_(a)xx4.526+6xxN_(a)xx13.6+6xxN_(a)xx12.75 rArr a=6 `moles |
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| 32. |
Radiation corresponding to the transition n=4 to n=2 in hydrogen atoms falls on a certain metal (work function=2.5 eV). The maximum kinetic energy of the photo-electrons will be:A. `0.55 eV`B. `2.55 eV`C. `4.45 eV`D. None of these |
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Answer» Correct Answer - 4 `E_(n)=-13.6/n^(2) eV, E_(2)=13.6/2^(2)` `E_(4)=-13.6/4^(2)eV//atom` DeltaE=E_(4)-E_(2)=2.55 eV` Absorbed energy = work function of metal `+K.E. 2.55=2.5+K.E., K.E.=0.05 eV` |
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| 33. |
Calculate the energy emitted when electron of 1.0 gm atom of Hydrogen undergo transition giving the spectrtal lines of lowest energy is visible region of its atomic spectra. Given that, `R_(H)`=`1.1xx10^(7) m^(-1)`,` c=3xx10^8m//sec`,` h=6.625xx10^(-34) Jsec`. |
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Answer» Correct Answer - `182.5 kJ` `E=13.6[1/n_(1)^(2)-1/n_(2)^(2)]eV//atom rArr E=13.6[(9-4)/36] eV//atom` `E=13.6[(9-4)/(36)]xx6.023xx10^(23)xx1.6xx10^(-19) rArr 182.5 kJ` |
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| 34. |
Calculate the wavelength of a photon emitted when an electron in `H-` atom maker a transition from `n=2` to `n=1` |
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Answer» `1/lambda=RZ^(2)[1/n_(1)^(2)-1/n_(2)^(2)]` `:. 1/lambda=R(1)^(2)[1/1^(2)-1/2^(2)]` `:. 1/lambda=(3R)/(4)` or `lambda=(4)/(3R)` |
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| 35. |
Calculate the two highest wavelength of the radiation emitted when hydrogen atoms make transition from higher state to `n = 2` |
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Answer» Correct Answer - `6561 Å, 4863 Å` (Approx) `lambda_(3 to 2)=12400/(DeltaE_(3 to 2))=12400/1.89=6561 Å, lambda_(4 to 2)=12400/(DeltaE_(4 to 2))=12400/2.55=4863 Å` |
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| 36. |
Calculate the frequency of light emitted in an electron transition from the sixth to the second orbit of a hydrogen atom. In what region of the specturm does this frequency occur? |
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Answer» Correct Answer - `v=7.3xx10^(14) Hz`, visible spectrum `DeltaE_(6 to 2)=hv rArr v=(3.022xx1.6xx10^(-19))/(6.625xx10^(-34))=7.3xx10^(14) Hz` This frequency lies in visible spectrum. |
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| 37. |
Atom A possesses higher values of packing fraction that atom B. The relative stabilities of A and B areA. A is more stable than BB. B is more stable than AC. A and B both are equally stableD. Stability does not depend on packing fraction |
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Answer» Correct Answer - B The atom which has lower value of packing fraction is stable |
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| 38. |
Atom A possesses higher values of packing fraction than atom B. The relative stabilities of A and B areA. Ais more stable than BB. B is more stable than AC. A and B both are equally stableD. Stability does not depend on packing fraction |
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Answer» Correct Answer - b The atom which has lower value of packing fraction is more stable. |
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| 39. |
The negative value of packing fraction indicates that the isotope isA. UnstableB. Very stableC. ArtificalD. Stable |
| Answer» Correct Answer - D | |
| 40. |
The nuclear binding energy for Ar (39.962384 amu) is : (given mass of proton and neuron are 1.007825 amu and 1.008665amu respectively)A. 343.62 MeVB. 0.369096 MeVC. 931 MeVD. None of these |
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Answer» Correct Answer - A `._(18)Ar^(40)` Total no. of protons = 18 Total no. of neutrons = 22 Mass defect `= [m xx p + m xx n] = 39.962384` `= [1.007825 xx 18 + 1.008665 xx 22] - 39.962384` `=[18.14085 + 22.19063] - 39.962384 = 0.369` Binding energy = mass defect `xx 931` `= 0.369 xx 931 = 343.62 MeV` |
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| 41. |
Calculate the packing fraction of `Ar^(40)` (isotopic weight of `Ar = 39.96238)`. |
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Answer» Isotopic atomic weight of `Ar^(40) = 39.96238` Mass number of `Ar^(40) = 40.00` `:.` Packing fraction ` = (39.96238 - 40)/(40) xx 10^(4)` `= (-0.03762 xx 10^(4))/(40) = -9.405` |
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| 42. |
The actual atomic mass of `._(20)Ca^(40)` is 39.96259 amu. Find the binding energy for this nuclide, using 1.008665 amu for the mass of a neutron and 1.007825 amu for the mass of atomic hydrogen. Also calculate the binding energy per nucleon. |
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Answer» Mass of `._(20)Ca^(40)` atom = Mass of 20 atoms of hydrogen + Mass of 20 neutrons `= 20 xx 1.007825 + 20 xx 1.008665 = 40.32980` amu Acutal mass of `._(20)Ca^(40)` atom = 39.96259 amu (given) `:.` Mass defect `(Delta M) = 40.32980 - 39.96259` = 0.36259 amu For a mass defect of 0.36721 amu, energy released `= 931.5 xx 0.36721 MeV = 342.06 MeV` |
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| 43. |
Sulphur-35 (34.96903 amu) emits a `beta -` particle but no `gamma-` rays, the product is chlorine -35 (34.96885 amu). The maximum energy emitted by the `beta`- particle isA. 0.016767 MeVB. 1.6758 MeVC. 0.16758 MeVD. 16.758 MeV |
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Answer» Correct Answer - C Mass defect = mass of sulphur - mass of chlorine `= 34.96903 - 34.96885 = 0.00018g` Binding energy = mass defect `xx 931` `= 0.00018 xx 931` `= 0.1675 MeV` |
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| 44. |
In the nuclear reaction `._(12)Mg^(24) + ._(2)He^(4) = ._(0)n^(1) +` ? The product nucleus isA. `._(13)Al^(27)`B. `._(14)Si^(27)`C. `._(13)Al^(28)`D. `._(12)Mg^(25)` |
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Answer» Correct Answer - B `._(12)Mg^(24) + ._(2)He^(4) rarr ._(0)n^(1) + ._(14)Si^(27)` |
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| 45. |
Respresentation of following nuclear reactions are as shown below: `._(7)N^(14) + ._(2)alpha^(4) rarr ._(8)O^(17) + ._(1)p^(1), {._(7)N^(14) (alpha, p) ._(8)O^(17)}` `._(13)Al^(27) + ._(2)alpha^(4) rarr ._(15)P^(30) + ._(0)n^(1) , {._(13)A^(27) (alpha, n) ._(15)P^(30)}` Write the missing particles in representation given below. `._(13)Al^(27), ._(8)O^(17) (--), ._(7)N^(14), ._(15)P^(30)`. Also write the corresponding nuclear reaction. |
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Answer» The given reaction is `._(7)N^(14) +._(2)alpha^(4) rarr ._(8)O^(17) + ._(1)P^(1), {._(7)N^(14) (alpha,p)._(8)O^(17)}` …..(i) or `._(8)O^(17) + ._(1)P^(1) rarr ._(7)N^(14) rarr ._(7)N^(14) + ._(2)alpha^(4), {._(8)O^(17) (p,alpha)._(7)N^(14)}` …..(ii) and `._(13)Al^(27) + ._(2)alpha^(4) rarr ._(15)P^(30) + ._(0)n^(1), {._(13)Al^(27) (alpha, n) ._(15)P^(30)}`....(iii) On adding Eqs. (ii) and (iii), we get `._(13)Al^(27) + ._(8)O^(17) + ._(1)p^(1) rarr ._(15)P^(30) + ._(7)N^(14) + ._(0)n^(1)` and it can be represented as `._(13)Al^(27), ._(8)O^(17) (p, n) ._(15)P^(30), ._(7)N^(14)` So, missing particles in representation are `p` and `n`. |
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| 46. |
`._(18)Ar^(40), ._(20)Ca^(40) and ._(19)K^(40)` areA. IsomersB. IsotopesC. IsobarsD. Isotones |
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Answer» Correct Answer - C Atoms of different elements having different atomic no.but same mass no. are called isobars |
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| 47. |
A radioactive nuclide emits `gamma`-rays due to a. K-electron capture b. Nuclear transition from higher to lower energy c. Presence of greater number of neutrons than protons d. Presence of greater of protons than neutronsA. a. K-electron captureB. b. Nuclear transition from higher to lower energyC. c. Presence of greater number of neutrons than protonsD. d. Presence of greater of protons than neutrons |
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Answer» Correct Answer - b. Nuclear transition from higher to lower energy b. After `alpha-,beta`-emission nucleus goes to excited state, when it returns to normal state, emission of `gamma`-radiations takes place. |
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| 48. |
Which of the following nuclear change is incorrect?a)`._(20)Ca^(40) + ._(0)n^(1) rarr ._(19)K^(40) + ._(1)H^(1)`b)`._(12)Mg^(24) + alpha rarr ._(14)Si^(27) + ._(0)n^(1)`c)`._(48)Cd^(113) + ._(0)n^(1) rarr ._(48)Cd^(112) + ._(-1)e^(0)`d)`._(20)Co^(43) + alpha rarr ._(21)Si^(46) + ._(1)H^(1)`A. `._(20)Ca^(40) + ._(0)n^(1) to ._(19)K^(40) +._(1)H^(1)`B. `._(12)Mg^(24) + ._(2)He^(4) to ._(14)Si^(27) + ._(0)n^(1)`C. `._(48)Cd^(113) + ._(0)n^(1) to ._(48)Cd^(112) + ._(-1)e^(0)`D. `._(20)Co^(43)+ ._(2)He^(4) to ._(21)Sc^(46) + ._(1)H^(1)` |
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Answer» Correct Answer - c A nuclear reaction must be balanced in terms of atomic numbers and mass number. Hence, following reaction is unbalanced. `._(48)Cd^(113) + ._(0)n^(1) to ._(48)Cd^(112) + ._(-1)e^(0)` |
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| 49. |
A heavier element containously emits `alpha`-and `beta`-particles. The finally stable element may belong to: a. 14th group b. 16th group c. 10th group d. 12 th groupA. a. 14th groupB. b. 16th groupC. c. 10th groupD. d. 12 th group |
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Answer» Correct Answer - a. 14th group a. Neptunium series ends at `Bi` (15th group) and rest all series terminate at `Pb` (14th group) |
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| 50. |
For the nuclear reaction `"_(12)^(24)Mg + d rarr alpha + `? The missing nuclide isA. `._(11)^(22)Na`B. `._(11)^(23)Na`C. `._(12)^(23)Mg`D. `._(12)^(26)Mg` |
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Answer» Correct Answer - A `._(12)^(24)Mg + ._(1)D^(2) rarr ._(2)He^(4) + ._(11)^(22)Na` |
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