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Physics NEET MCQ
Atom And Nucleus Mcq
Quiz 8
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Q.1
) Paragraph The mass of a nucleus ZXA is less than the sum of the masses of (A-Z) number of neutronsand Z number of protons in the nucleus. The energy equivalent to the corresponding massdifference is known as the binding energy of the nucleus. A heavy nucleus of mass M canbreak into two light nuclei of masses m1 and m2 only if (m1 + m2) < M. Also two light nucleiof masses m3 and m4 can undergo complete fusion and form a heavy nucleus of mass M′ only if (m3 + m4) > M′. The masses of some neutral atoms are given in the table below : The correct statement is
1
H
1
1.007825 u
1
H
2
2.014102 u
1
H
3
3.016050u
2
He
4
4.002603u
3
Li
6
6.015123 u
3
Li
7
7.016004 u
30
Zn
70
69.925325u
34
Se
82
81.916709u
64
Gd
152
151.919803u
82
Pb
206
205.974455u
83
Bi
209
208.980388u
84
Po
210
209.982876u
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a) The nucleus 3Li6 can emit an alpha particle.
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b) The nucleus 84Po210 can emit a proton.
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c) Deuteron and alpha particle can undergo complete fusion.
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d) The nuclei 30Zn70 and 34Se82 can undergo complete fusion
Explanation
Option a M = 6.015123 m1 = 4.002603 m2 = 2.014102 Total of m1 and m2 = 6.016705 u Breaking possible if (m1 + m2) < M but (m1 + m2) > M so not possible Option b M=205.974455u m1 = 4.002603u m2 = 205.974455u Total of m1 and m2 = 209.988213 u (m1 + m2) > M so not possible Option c M’=6.015123 u m3 = 2.014102 u m4 = 4.002603u m3 + m4=6.016705u Fusion possible if (m3 + m4) < M’ But (m3 + m4) > M’ so not possible Option d M’=151.919803u m3 = 69.925325u m4 = 81.916709u m3 + m4 =151.842034 u As (m3 + m4) < M’ fusion possible Answer:(d)
Q.2
The kinetic energy (in keV) of the alpha particle, when the nucleus 84Po210 at restundergoes alpha decay, is
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a) 5319
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b) 5422
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c) 5707
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d) 5818
Explanation
M = 209.982876 m1 = 205.974455 m2 = 4.002603 m1 + m2 = 209.977058 M - (m1 + m2) = 0.005818 1 u = 932 MeV ∴ Energy released = 0.005818 × 932 = 5.422MeV=5422 KeV This energy will go to both the nuclei as kinetic energy From law of conservation of momentum m1v1=m2v2 206v1 = 4v2 K.E of alpha = 5318.72keV ≈ 5319 keV Answer:(a)
Q.3
Match List I of the nuclear processes with List II containing parent nucleus and one of the end products of each process and then select the correct answer using the codes given below the lists:
List I
List II
P. Alpha decay
8
O
15
→
7
N
15
+ ....
Q. β+ decay
92
U
238
→
90
Th
234
+ ....
R. Fission
83
Bi
185
→
82
Pb
184
+ ....
S. Proton emission
94
Pu
239
→
57
La
140
+ ...
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a) P → 4 Q → 2 R → 1 S → 3
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b) P → 1 Q → 3 R → 2 S → 4
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c) P → 2 Q → 1 R → 4 S → 3
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d) P → 4 Q → 3 R → 2 S→ 1
Explanation
Answer:(c)
Q.4
)If λCu is the wavelength of Kα X-ray line of copper (atomic number 29) and λMo is the wavelength of the Kα X-ray line of molybdenum (atomic number 42), then the ratio λCu /λMo is close to
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a) 1.99
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b) 2.14
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c) 0.50
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d) 0.48
Explanation
Answer:(b)
Q.5
A fission reaction is given by 92U236→ 54Xe140 + 38Sr94+x+y,where x and y are two particles. Considering 92U236 to be at rest, the kinetic energies of the products are denoted by KXe, KSr, Kx (2 MeV) and Ky (2 MeV), respectively. Let the binding energies per nucleon of 92U236,54Xe140,38Sr94 be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is (are)
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a) x = n, y = n, KSr = 129 MeV, KXe = 86 MeV
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b) x = p, y = e, KSr = 129 MeV, KXe = 86 MeV
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c) x = p, y = n, KSr = 129 MeV, KXe = 86 MeV
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d) x = n, y = n, KSr = 86 MeV, KXe = 129 MeV
Explanation
Q Value of the reaction Q = 94 × 8.5 + 140 × 8.5 - 236 × 7.5 Q = 219 MeV X and Y share 4 MeV together ∴ remaining 215 MeV will be shared between Xe and Sr nucleus. On the basis of law of conservation of charge and mass energy Only option “a” is correct. Answer:(a)
Q.6
Highly excited states for hydrogen-like atoms (also called Rydberg states) with nuclear charge Ze are defined by their principal quantum number n, where n >>Which of the following statement(s) is(are) true? [ IIT Advance 2016]
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a) Relative change in the radii of two consecutive orbitals does not depend on Z
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b) Relative change in the radii of two consecutive orbitals varies as 1/n
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c) Relative change in the energy of two consecutive orbitals varies as 1/n3
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d) Relative change in the angular momenta of two consecutive orbitals varies as 1/n
Explanation
Relative change in radii Option a and option b correct Angular momentum L ∝ n Option d correct Energy E∝1/n2 ΔE ∝ 1/n3 Option c wrong Answer:(a, b, d)
Q.7
P is the probability of finding the 1s electron of hydrogen atom in a spherical shell of infinitesimal thickness, dr, at a distance r from the nucleus. The volume of this shell is 4πr2 dr. The qualitative sketch of the dependence of P on r is… [ IIT Advance 2016]
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a)
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b)
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c)
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d)
Explanation
Option c is the standard distribution Answer:(c)
Q.8
The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by The measured masses of the neutron , 1H1, 7N15 and 8O15, are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u, respectively. Given that the radii of both the and nuclei are same, 1 u = 931.5 MeV/c2 (c is the speed of light) and e2/(4πε0 ) = 1.44 MeVfm Assuming that the difference between the binding energies of 7N15 and 8O15 is purely due to the electrostatic energy, the radius of either of the nuclei is …[IIT Advance 2016] (1 fm = 10-15 m)
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a) 2.85 fm
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b) 3.03 fm
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c) 3.42 fm
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d) 3.80 fm
Explanation
Given that radius depends on number of protons = Binding energy Binding energy of 7N15 = mP7 +mn8 – mass of N Binding energy of 8O15 = mP8 +mn7 – mass of O ΔB.E. = Binding energy of 8O15 - Binding energy of 8O15 ΔB.E.= (mP8 +mn7 – mass of O - mP7 -mn8 + mass of N) 931.5MeV ΔB.E.= (mP - mn – mass of O +mass of N) 931.5MeV = (1.008665 u- 1.007825 u + 15.000109 u - 15.003065 u) 931.5MeV ΔB.E.= 0.003796 u × 931.5MeV R = 3.42 fm Answer:(c)
Q.9
An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use? [IIT Advance 2016]
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a) 64
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b) 90
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c) 108
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d) 120
Explanation
I=I0 e(-λt) 64=eλt ln64=λt t=8×18=108 days Answer:(c)
0 h : 0 m : 1 s
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