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Physics NEET MCQ
Atom And Nucleus Mcq
Quiz 5
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Q.1
9) which of the following can not be emitted in radioactive decay of the substance?
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a) Helium-nucleus
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b) Electrons
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c)Neutrions
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d)Proton.
Explanation
Answer:(d)
Q.2
0) If the radius of 27 13 Al nucleus is 3.6 fm the radius of 125 52 Te nucleus is nearly equal to
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a) 8 fm
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b) 6 fm
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c) 4 fm
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d) 5 fm
Explanation
By using the formula, Answer: (b)
Q.3
) which of the following atom has the lowest ionization potentical?
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a) 147 N
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b) 4018 Ar
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c)13355 Cs
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d)168 O
Explanation
55 Cs133 a nucleus having highest Z number, the outer most electron are less inbinding with nuclera. That means lowest binding energyAnswer: (c)
Q.4
) If the binding energy of electron in a hydrogen atom is 13.6 eV, the energy required to remove the elecron form the first state of Li2+ is.
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a) 13.6 eV
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b) 30.6 eV
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c) 122.4 eV
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d)3.4 eV
Explanation
Binding energy=13.6 z2 / 2 For Li2+ , Z=3, n=2 first exited stateAnswer: (b)
Q.5
) The ionigation Potential of hydrogen atom is 13.6 eV. An electron in the ground state absords Photon of energy 12.75 eV. How many dirrerent spectral lines can one expect when electron make a down ward transition
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a) 1
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b)2
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c)6
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d)4
Explanation
Answer:(d)
Q.6
) A radio-active nucleus A Z X emits 3α -particles and 2 Positrions. the ratio of number of neuleuons to that of Protons in the final nucleus will be
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a)( A - Z - 8 ) / Z - 4
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b) ( A + Z - 12 ) / Z - 4
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c) (A - Z - 4 ) / Z - 8
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d) ( A - Z - 4 ) / Z - 2
Explanation
Answer: (c)
Q.7
) An α-particle of energy ½ mv2 bombards by a heavy nuclear target of charge .Thenthe distance of closest approach for the alpha nucleus will be Proportional to
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a)1/Z3
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b) 1/ v4
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c)1 / m
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d)1 / v2
Explanation
Answer: (c)
Q.8
6) when 73 Li nucler are bombarded by Proton and the resultant nuclei are 84 Be, then emitted particle will be
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a) neutron
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b) gamma
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c) alpha
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d)Beta
Explanation
Answer: (b)
Q.9
) starting with a sample of Puer cu-66, 7/8 of its decays into Zn in 15 min. The corresponding half-life is
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a)5 min
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b)7.5 min
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c)10 min
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d) 15 min
Explanation
Let initial amount of Cu be N. Fraction decayed in 1st half life = ½ N Fraction decayed in 2nd half lives = ¼ N Fraction decayed in 3 half lives = ⅛ N Total amount decayed in 3 half lives = ½ N + ¼ N + ⅛ N = (7/8) N ∴ 3 half lives = 15 minutes ∴ Half life of the Cu is 5 minutes. Answer:(a)
Q.10
) An α -particle of energy 5 MeV is scattered though 180 by a fixed uranium nucleus. The distance of the closest approach nucleus The distance of the closest approach is of the order of
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a) 10-8 cm
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b) 10-12 cm
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c) 10-10 cm
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d) 10-15 cm
Explanation
Answer: (b)
Q.11
The binding energy Per nucleon of deutron (2 1H) and Lielium nucleus (4 2He ) 1.1 MeV and 7.0 MeV.respectively. If two beutron nucler react to form a singlehelium nucleus, the energy released is
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a) 23.6 MeV
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b) 26.9 MeV
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c)13.9 MeV
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d)19.2 MeV
Explanation
Answer: (a)
Q.12
) The nucleus at rest disintegrate into two nuclear parts which have their velocities in the ratio 2:1 The ratio of their nuclar sizes will be
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a) 3√2 :1
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b) 1 : 3 √2
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c) 3 : 1
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d)1 : 3
Explanation
According to cosveration momentum, Answer: (b)
Q.13
) A radiation of energy E falls normally on a Pertect reflecting surface. The momentum transferred to the surtace is.
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a) E / c
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b) 2E / c
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c)E / c2
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d)Ec
Explanation
Here, surface is perfacet reflector momentum of incident radiation is E/C momentum of reflected rediation is - E/C, change in momentum=- E /c - E / c=-2E /c ∴ Momentum transtered to the surface=2E / c Answer:(b)
Q.14
) In the following nuclear fusion reaction the repalsive potential energy between the two fusing nucler is 7.7× 10-14 J The Temperature to which the gas must be heated is nearly (Boltzman constmt K=1.38 × 10 -23 J / K )
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a) 103K
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b) 105K
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c) 107K
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d) 109K
Explanation
Both the nucleus have same charge hence repel each other, to overcome repulsion and undergo nuclear fussion, equal amount of kinetic energy must be provided in the form of heat energy Kinetic energy of gas at a given temperature is given by following formula E=(3/2)KT By substituting values 7.7× 10-14 = (3/2)1.38 × 10 -23 Answer: (d)
Q.15
) If the binding energy Per nucleon in 73 Li and 42 He nucler is 5.6 MeV and 7.06 MeV respectively, then in the reaction (P here retrent Proton)energy of Protpn must be
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a) 1.46 MeV
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b) 39.2 MeV
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c) 17.28 MeV
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d)17.28 MeV
Explanation
Energy of proton=4 (7.06)2 - 7(5.6)=2×28.24 -39.2=56.48-39.2=17.28Answer: (d)
Q.16
) In gamma ray emission form a nucleus
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a) there is no change in the proton-number and neutron number
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b) Both the number are changes
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c)only Proton number change
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d)only neutron number change
Explanation
Answer: (a)
Q.17
) The half life time of a radidactive elements of x is the same as the mean life of another radioactive element Y. Initially they have same number of atoms, then
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a) y will decay faster than x
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b) x will decay faster than y
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c) x and y will decay at the same rate at all time
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d)x and y will decay at the same rate intially
Explanation
Half life = 0.693 × mean life It means Half life time > mean life time Given Half life of X = mean life of Y ⇒ Half life of Y is less than Half life of X Smaller the half life faster it will decay Thiu Y will decay fast Answer:(a)
Q.18
) which of the following transition in hydrogen atoms emits Photon of highest frequengy?
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a) n=2 to n=6
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b) n=1 to n=2
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c) n=2 to n=1
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d) n=6 to n=2
Explanation
n=2 to n=6 absorbs photon, n=9 to n=2 absorbs photon,n=6 to n=2 emission of photon, n=2 to n=1 emission of photon. Answer: (c)
Q.19
) An electron Passing through a Potential diffencne of 4.9 v colides with a mercury atom and trnasfer it to the first excited state what is trnasfer it to the first excited state. what is the wave length of Photon corresponding to the franition of mercury atom to its normal state.
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a) 2050A
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b) 2935A
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c)2525A
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d)2240A
Explanation
Energy of electron=E=ve E=4.9 × 1.6 × 10 -19 J E=hc / λ ∴ λ=hc / EAnswer: (c)
Q.20
) The binding energy per nucleon for the Parent nucleus is E1 and that for the daughter nuclei is E 2 then
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a) E 1 < E 2
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b) E 1 > E 2
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c) E 1=E 2
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d) E 1=2 E 2
Explanation
Answer: (a)
Q.21
) 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
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c)1.744 × 10-19
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d)1.168 × 10-19
Explanation
K max=hf - Φ ∴ hf=K max + Φ Answer:(a)
Q.22
) 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. Ionization Potential of hydrogen is 13.6 v and the mass of hydrogen atom is 1.67 × 10-27 kg. The quantum number of the two levels in the emission
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a) n=1 , n=3
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b) n=2 , n=4
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c) n=1 , n=4
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d) n=3 , n=4
Explanation
When radiation falls on surface, it energy is utilized to overcome bonding energy and remaing as kinetic energy It can be given by mathamatical equation as Incidant radiation energy (hν) = Work function (φ) + Kinetic energy Given Work function = 1.82 ev and Photoelectron energy = 0.73 ev Sunstituting above values in eqation we get Energy of incidant radiation hν = 1.82 +.73 =2.25 ev Energy of Hydrogen atom orbit is 13.6/n2 Clculate values of energy for different energy elevel and subtract Corresponding energy level for 2.55 eV is n=2 and n=4Answer: (b)
Q.23
) 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. In this transition change in the angular momentum of electron is (where h isPlank constanst )
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a)h / 2π
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b) h / π
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c)2h / π
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d)3h / 2π
Explanation
change in Angular momentum=(4h / 2π) - ( 2h / 2π)=h / π Answer: (b)
Q.24
)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 recoil speed of emitting atom caussing that is at lest before the transitionis of the order of
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a) 1 cm /s
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b) 102 m / s
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c)104 m / s
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d)1 m /s
Explanation
According to conservation of momentum, momentum of proton=moentum of recoil atom ∴ h / λ=m &mew μ=h / mλ=E / mcAnswer: (d)
Q.25
) A and B are two radioactive substane whose half lives are 1 and 2 years respectively. Initially 10 g of A and 1 g of B is taken. The time after which they will have same quantity remaining is
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a) 3.6 years
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b) 7 years
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c)6.6 years
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d)5 years
Explanation
Answer:(c)
Q.26
) which of these is a fusion reuction
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a) 31H + 21H=42 + 10n
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b) 127C → 126 C + β + + γ
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c) 23892 → 29882 Pb + 8( 42He) + 6( 10e)
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d) none of these
Explanation
Answer: (a)
Q.27
) The activity of a sample of a radio- active material is at time t1 and A2 at time t 2 (where t2 > t1) if T its mean life is then
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a)A1t1 A2t2
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b)
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c)A2 - A1=t2 - t1
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d)A2=A1 e(t2 / t1)T
Explanation
Answer: (b)
Q.28
6) The energy released by the fission of one unanium atom is 200 MeV. The number of fission Per second required to Produce 3.2 w of Power is
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a) 1010
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b) 107
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c) 1012
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d) 1011
Explanation
Energy released per uranium atom=200 MeV Answer: (d)
Q.29
) In the following disintegration series 92 U 238 → α→ X → -β → z Y A The value of Z and A respectively will be
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a)90 , 234
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b) 92 , 236
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c)88 , 234
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d)91 , 234
Explanation
92 U 238 → 90 Tb 234 → 91 Y 234 When one α emiits it decreases atomic number by 2 and mass number by 4 When -β particle emmits it changes one Neutron to proton N → P + (-β) Answer:(d)
Q.30
) If 92 U 238 undergoes sucesively 8α decays and 6β decays then resulting nucleus is
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a) Pb 206
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b) Pb 208
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c) Pb 214
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d) None of these
Explanation
92 U 238 → 8 α → 76 X 204 → 6β → 82 Y 206Answer: (a)
0 h : 0 m : 1 s
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