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Physics NEET MCQ
Quiz 7
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Q.1
) Excited hydrogen atom emits a photon of wave length in return to the groundstate The quantum number n of excited state is given by
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a)
0%
b)
0%
c)
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d)λR(R-1)
Explanation
Answer:(a)
Q.2
) The radius of hydrogen atom in the first excited level is{AIIMS 1998}
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a) Twice
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b) four times
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c) same
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d) half
Explanation
Radius of H-atom varies as a square of n So for excitation from n=1 to n=2, radius becomes 4 times.Answer: (b)
Q.3
Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is … [NEET 2013]
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a) 5/27
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b) 3/23
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c) 7/29
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d) 9/31
Explanation
For longest wave length Layman series Balmer series Answer:(a)
Q.4
The half life of a radioactive isotope ‘X’ is 20 years.It decays to another element ‘Y’ which is stable. Thetwo elements ‘X’ and ‘Y’ were found to be in theratio 1 : 7 in a sample of a given rock. The age ofthe rock is estimated to be…[NEET 2013]
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a) 40 years
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b) 60 years
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c) 80 years
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d) 100 years
Explanation
If initial is N0 = 8 then balance N =1 After 20years = 4 Next 20 years = 2 Next 20 years =1 Thus after 60 years ratio of X:Y= 1:7 Alternate method 3 half lives, T = 3 × 20 = 60 years Answer:(b)
Q.5
A certain mass of Hydrogen is changed to Heliumby the process of fusion. The mass defect in fusionreaction is 0.02866 u. The energy liberated per u is…[NEET 2013] (given 1 u = 931 MeV)
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a) 2.67 MeV
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b) 26.7 MeV
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c) 6.675 MeV
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d) 13.35 MeV
Explanation
Answer:(c)
Q.6
The binding energy per nucleon of 3Li7 and 2He4 nuclei are 5.60 MeV and 7.06 MeV, Respectively. In the nuclear reaction the value of energy Q released ….
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a) 8.4 MeV
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b) 17.3 MeV
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c) 19.6 MeV
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d) – 2.4 MeV
Explanation
Q = 2(4×7.06) - 7×5.6 Q = 56.48 – 39.2 = 17.28 MeV Answer:(b)
Q.7
A radio isotope 'X' with a half life 1.4 × 109years decays to 'Y'which is stable.Asampleof the rock froma cavewas found to contain'X' and 'Y' in the ratio 1:The age of therock is[ AIPMT 2014]
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a) 4.20 × 109 years
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b) 8.40 × 109 years
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c) 1.96 × 109 years
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d) 3.92 × 109 years
Explanation
If initial is N0 = 8 then balance N =1 3 half lives, T = 3 × 1.4 × 109 = 4.2 × 109 years Answer:(a)
Q.8
Hydrogen atom is ground state is excited by amonochromatic radiation of λ=975 Å.Number of spectral lines in the resulting spectrum emitted will be
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a) 6
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b) 10
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c) 3
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d) 2
Explanation
Energy of incident photon is Energy=12.74ev Excited state of hydrogen atom, electron have energy = -13.6 +12.75 = -0.85 eV Or n=4 thus electron goes to n=4 Now possible numbers of transitions are given by For n =4 number of transitions are 6 Answer:(a)
Q.9
If radius of 13Al27 nucleus is taken to be RAl, then the radius of 53Te125 nucleus is nearly … [AIPMT 2015]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Form formula R = A(Z)1/3 RAl = A(27)(1/3) = 3A…(i) Now R=A(125)(1/3) = 5A…(ii) From (i) and (ii) R = (5RAl)/3 Answer:(c)
Q.10
In the spectrum of hydrogen, the ratio of the longestwavelength in the Lyman series to the longest wavelength in the Balmer series is : ..[ Re-AIPMT 2015]
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a) 5/27
0%
b) 4/9
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c) 9/4
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d) 27/5
Explanation
For Lyman series For Balmer series Answer:(a)
Q.11
A nucleus of uranium decays at rest into nuclei ofthorium and helium. Then :-
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a) The helium nucleus has less kinetic energy thanthe thorium nucleus
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b) The helium has more kinetic energy than thethorium nucleus.
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c) The helium nucleus has less momentum than thethorium nucleus.
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d) The helium nucleus has more momentum thanthe thorium nucleus.
Explanation
According to law of conservation of linear momentum Momentum of thorium = momentum of helium Mv1 = mv2 Since momentum of both the particles is same energy ∝ (1/m) Since mass of helium is less than mass of thorium Kinetic energy of Helium > Kinetic energy of thorium Answer:(b)
Q.12
When an α-particle of mass 'm' moving with velocity'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleusdepends on m as :
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a) 1/m
0%
b) 1/√m
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d) m
0%
c) 1/m²
Explanation
At closest distance of approach, the kinetic energyof the particle will convert completely intoelectrostatic potential. Thus d ∝ 1/m Answer:(a)
Q.13
Given the value of Rydberg constant is 107m–1, thewave number of the last line of the Balmer series in hydrogen spectrum will be :-
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a) 0.025 × 104 m–1
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b) 0.5 × 107 m–1
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c) 0.25 × 107 m–1
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d) 2.5 × 107 m–1
Explanation
For last line electron to migrate from infinite to n = 2 in case of Balmaer series. For hydrogen atom Z=1 Wave number = 1/λ Answer:(c)
Q.14
If an electron in a hydrogen atom jumps from the3rd orbit to the 2nd orbit, it emits a photon ofwavelength λ. When it jumps from the 4th orbit tothe 3rd orbit, the corresponding wavelength of thephoton will be …[NEET II -2016]
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a) 20λ /7
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b) 20λ /13
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c) 16λ /25
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d) 9λ /16
Explanation
Transition: 3 → 2 ⇒ Wavelength λ, Transition: 4 → 3 ⇒ Wavelength λ’= ? Answer:(a)
Q.15
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is …[ NEET II – 2016]
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a) 45
0%
b) 60
0%
c) 15
0%
d) 30
Explanation
Initial decay % = 40% ∴ 60% material is un-decay Now half life is 30 minutes In 30 min substance will decay 50% of 60% of un-decay to give 30% decay and 30% undecayed and total decay =30%+40% = 70% decay In next 30 min substance will decay by 50% of 30% will give 15% decay, and undecay = 15% Thus at the end of secind half life decay is 85% and un-decay is 15% Therefore 2 half life taken from between 40% decay and 85% decay t= 2t1/2 = 2×30 = 60 min Answer:(b)
Q.16
Radioactive material 'A' has decay constant '8λ' andmaterial 'B' has decay constant 'λ'. Initially they havesame number of nuclei. After what time, the ratio ofnumber of nuclei of material 'A' to that 'B' will be 1/e
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a) ) 1/λ
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b) 1/7λ
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c) 1/8λ
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d) 1/9λ
Explanation
N=N0 e(-kt) 7λt= 1 , ∴ t=1/7λ Answer:(b)
Q.17
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is
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a) 2
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b) 1
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c) 4
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d) 0.5
Explanation
For last Balmer series For last Lyman series Answer:(c)
Q.18
In a hydrogen like atom electron makes transition from an energy level with quantum number nto another with quantum number (n − 1). If n >> 1, the frequency of radiation emitted is proportional to …[ IIT Mains 2013]
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a) 1/n
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c) 1/n3/2
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b) 1/n²
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d) 1/n³
Explanation
Answer:(d)
Q.19
The radius of the orbit of an electron in a Hydrogen-like atom is 4.5 a0, where a0 is the Bohr radius. Its orbital angular momentum is 3h/2π. It is given that h is Planck constant and R is Rydberg constant. The possible wavelength(s), when the atom de-excites, is (are)
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a) 9/32R
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b) 9/16R
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c) 9/5R
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d) 4/3R
Explanation
From Bohr’s second hypothesis n = 3 Z = 2 Three transitions are possible from 3 → 2 and 3 → 1 also 2 → 1 from (i) 3 → 2 3 → 1 2 → 1 Answer:(a, c)
Q.20
) 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.21
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.22
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.23
)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.24
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
0%
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.25
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.26
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)
0%
c)
0%
d)
Explanation
Option c is the standard distribution Answer:(c)
Q.27
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.28
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)
Q.29
The radius of hydrogen atom in its ground state is 5.3×10⁻¹¹ m. After collision with an electron it is found o have a radius of 21.2×10⁻¹¹ m. What is the principal quantum number n of the final state of the atom.. [ CBSE-PMT 1994]
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a) n=4
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b) n=2
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c)n=16
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d)n=3
Explanation
r ∝ n2 ∴ radius of final state / radius of initial state=n2 Answer: (b)
Q.30
) Light form the discharge tube containing hydrogen atom falls on the surface of a Piece of sodium. The kinetic energy of the fastest photo electrons emitted form sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionigation Potential of hydrogen is 13.6 v and the mass of hydrogen atom is 1.67× 10⁻²⁷ kg. The energy of Photon causing the Photo electeic emission is
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a) 4.08 × 10⁻¹⁹ J
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b) 2.912 × 10⁻¹⁹
0%
c)1.744 × 10⁻¹⁹
0%
d)1.168 × 10⁻¹⁹
Explanation
K max=hf - Φ ∴ hf=K max + Φ Answer:(a)
0 h : 0 m : 1 s
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