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Physics NEET MCQ
Atom And Nucleus Mcq
Quiz 7
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Q.1
) A heavy nuecleus at least breaks into two frigments which fly off with velocities in the ratio 8:1 The ratio of radil of the frigments is
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a)1 : 2
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b) 4 : 1
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c)1 : 4
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d)2 : 1
Explanation
Answer: (a)
Q.2
) The binding energy Per nucleon of 8 O 16 is 7.97 MeV and that of 8 O 17 is 7.75 MeV The energy (in-MeV) required to remove a neutrom from 8 O 17 is
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a) 3.65
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b) 7.86
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c)3.52
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d)4.23
Explanation
Answer: (d)
Q.3
) The shape of the graph lnI → t is
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a)stright Line
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b)Parabolic curve
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c) Hyberbole curve
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d) random shape curve
Explanation
Answer:(a)
Q.4
) The Probability of survival of a radioactive nucleus for one mean life time is
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a) 1 - 1 /e
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b) 1 /e
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c) 2 / e
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d) 3 / e
Explanation
Answer: (b)
Q.5
) As the electron in Bohr is orbit of hydrogen atom passes from state n=2 to n=1, the K.E. and Potential energy changes as
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a)Two fold , also two fold
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b) four fold , two fold
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c)four fold , also four fold
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d) two fold , four fold
Explanation
kn = kinetic energy ∝(1 / n2) ∴ (k1 / k2) = (2 / 1)2 = (4 / 1) similarly potential energy un ∝ (1 / n2) ∴ (u1 / u2) = (2 / 1)2 = (4 / 1) i.e. KE and PE both changes as four fold. s Answer: (c)
Q.6
)The wave lenght of the first line of Lyman series for hydrogen atom is equal to that of hydrogen atom is equal to that of second line of Balmar series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
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a) 1
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b) 2
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c)3
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d)4
Explanation
Answer: (b)
Q.7
) the nuclear decay below z X A → z+1 Y A → z-1 B A-4 → z-1 B A-4
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a)β , α γ
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b)α,β , γ
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c)γ , β , α
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d)β , γ α
Explanation
Answer:(a)
Q.8
) A nucleus n X m emmits one α -particle and two β particle. The resulting nucleus is
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a) n-2 Y m-4
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b) n Y m-6
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c) n Y m-4
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d) n-4 Y m-6
Explanation
Answer: (c)
Q.9
) At a certain time, a radio active sample contains 2× 1020 atoms and disintegration rate is 3× 1010 atom per sec. when 2 × 1015 atoms are Left to decay its disintegationrate will be
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a)3 × 105atoms / s
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b) 3 × 1010 atoms / s
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c)6.6 × 101atoms / s
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d)2 × 102 atoms / s
Explanation
Answer: (a)
Q.10
) The radius of Ge nuclide is measured to be twice the radius of 4 Be 9 The number of nucleons in Ge are
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a) 72
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b) 78
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c)65
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d)80
Explanation
As Radius ∝ A1/3 Answer: (a)
Q.11
) 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)
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b)
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c)
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d)λR(R-1)
Explanation
Answer:(a)
Q.12
) 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.13
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.14
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.15
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.16
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.17
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.18
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.19
If radius of 13Al27 nucleus is taken to be RAl, then the radius of 53Te125 nucleus is nearly … [AIPMT 2015]
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a)
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b)
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c)
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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.20
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
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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.21
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.22
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
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b) 1/√m
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c) 1/m2
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d) m
Explanation
At closest distance of approach, the kinetic energyof the particle will convert completely intoelectrostatic potential. Thus d ∝ 1/m Answer:(a)
Q.23
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.24
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.25
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
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b) 60
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c) 15
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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.26
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.27
Suppose the charge of a proton and an electron differslightly. One of them is–e, the other is (e + Δe). Ifthe net of electrostatic force and gravitational forcebetween two hydrogen atoms placed at a distanced (much greater than atomic size) apart is zero,then Δe is of the order of [Given mass of hydrogen mh = 1.67 × 10–27 kg]
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a) a) 10–20 C
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b) 10–23 C
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c) 10–37 C
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d) 10–47 C
Explanation
FE = FG Charge on each hydrogen atom = -e+e+Δe = Δe Answer:(c)
Q.28
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.29
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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b) 1/n2
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c) 1/n3/2
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d) 1/n3
Explanation
Answer:(d)
Q.30
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)
0 h : 0 m : 1 s
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