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Nuclei MCQs for NEET — Physics Questions with Answers

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For uranium nucleus how does its mass vary with volume

$m \propto V$

$m \propto 1 / V$

$m \propto \sqrt{V}$

$m \propto V^{2}$

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Explanation

(a) Since nuclear density is constant hence mass ∝ volume.

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The rest mass of an electron as well as that of positron is 0.51 MeV. When an electron and positron are annihilated, they produce gamma-rays of wavelength(s)-

0.012 Å

0.024 Å

0.012 Å to $\infty$

0.024 Å to $\infty$

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Explanation

(a) Since electron and positron combine-
$λ = \frac{hc}{E_{Total}} = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{(0.51 + 0.51) \times 10^{6} \times 1.6 \times 10^{- 19}} = 1.21 \times 10^{- 12} m = 0.012 \overset{0}{A}$

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In the nuclear fusion reaction ${}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}{He} + n$ given that the repulsive potential energy between the two nuclei is $- 7.7 \times 10^{- 14} J$ , the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s constant $k = 1.38 \times 10^{- 23} J / K$ )

$10^{9} K$

$10^{7} K$

$10^{5} K$

$10^{3} K$

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Explanation

(a) Kinetic energy of the molecules of a gas at a temp. T is $\frac{3}{2} kT$
$∴$ To initiate the reaction $\frac{3}{2} kT = 7.7 \times 10^{- 14} J$

$\Rightarrow \frac{3}{2} \times 1.38 \times 10^{- 23} T = 7.7 \times 10^{- 14} \Rightarrow T = 3.7 \times 10^{9} K$
.

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A nucleus with mass number 220 initially at rest emits an $α$ -particle. If the Q value of the reaction is 5.5 MeV, calculate the kinetic energy of the $α$ -particle

4.4 MeV

5.4 MeV

5.6 MeV

6.5 MeV

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Explanation

The Q value of a nuclear reaction is the difference between the sum of the masses of the reactants and the sum of the masses of the products. In this case, the Q value is positive, indicating an exothermic reaction. The kinetic energy of the emitted alpha particle is equal to the Q value.

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The half life of radioactive Radon is 3.8 days. The time at the end of which 1/20 th of the Radon sample will remain undecayed is
(Given $\log_{10} e = 0.4343$ )

3.8 days

16.5 days

33 days

76 days

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Explanation

(b) By the formula $N = N_{0} e^{- λt}$
Given $\frac{N}{N_{0}} = \frac{1}{20}$ and $λ = \frac{0.6931}{3.8} \Rightarrow 20 = e^{\frac{0.6931 \times t}{3.8}}$
Taking log of both sides
or $\log 20 = \frac{0.6931 \times t}{3.8} \log_{10} e$
or $1.3010 = \frac{0.6931 \times t \times 0.4343}{3.8} \Rightarrow t = 16.5 days$

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If 10% of a radioactive material decays in 5 days, then the amount of original material left after 20 days is approximately

60%

65%

70%

75%

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Explanation

(b) $N = N_{0} e^{- λt}$

$∴ 0.9 N_{0} = N_{0} e^{- λ \times 5} \Rightarrow 5 λ = \log_{e} \frac{1}{0.9}$ ....(i)

and ${xN}_{0} = N_{0} e^{- λ \times 20} \Rightarrow 20 λ = \log_{e} (\frac{1}{x})$ ....(ii)

Dividing (i) and (ii) , we get

$\frac{1}{4} = \frac{\log_{e} (1 / 0.9)}{\log_{e} (1 / x)} = \frac{\log_{10} (1 / 0.9)}{\log_{10} (1 / x)} = \frac{\log_{10} 0.9}{\log_{10} x} \Rightarrow \log_{10} x = 4 \log_{10} 0.9 \Rightarrow x = 0.658 = 65.8 %$

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A radioactive isotope X with a half-life of $1.37 \times 10^{9}$ years decays to Y which is stable. A sample of rock from the moon was found to contain both the elements X and Y which were in the ratio of 1 : 7. The age of the rock is

$1.96 \times 10^{8}$ years

$3.85 \times 10^{9}$ years

$4.11 \times 10^{9}$ years

$9.59 \times 10^{9}$ years

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Explanation

(c) If in the rock there is no Y element, then the time taken by element X to reduce to $\frac{1}{8} th$ the initial value will be equal to $\frac{1}{8} = {(\frac{1}{2})}^{n}$ or n =3
Therefore, from the beginning three half life time is spent. Hence the age of the rock is

$= 3 \times 1.37 \times 10^{9} = 4.11 \times 10^{9} years$
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From a newly formed radioactive substance (Half life 2 hours), the intensity of radiation is 64 times the permissible safe level. The minimum time after which work can be done safely from this source is

6 hours

12 hours

24 hours

128 hours

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Explanation

(b) $\frac{N}{N_{0}} = {(\frac{1}{2})}^{n} \Rightarrow \frac{1}{64} = {(\frac{1}{2})}^{6} = {(\frac{1}{2})}^{n} \Rightarrow n = 6$

After 6 half lives intensity emitted will be safe.
$∴$ Total time taken = $6 \times 2 = 12 hrs$

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The half life of radium is 1620 years and its atomic weight is 226 kgm per kilomol. The number of atoms that will decay from its 1 gm sample per second will be

$3.61 \times 10^{10}$

$3.6 \times 10^{12}$

$3.11 \times 10^{15}$

$31.1 \times 10^{15}$

(Avogardo's number N = $6.02 \times 10^{26}$ atpm/kilomol)

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Explanation

(a)

$\frac{dN}{dt} = λN; λ = \frac{0.6931}{t_{1 / 2}} = \frac{0.6931}{1620 \times 365 \times 24 \times 60 \times 60} N = \frac{6.023 \times 10^{23}}{226}$

$∴ \frac{dN}{dt} = \frac{0.6931 \times 6.023 \times 10^{23}}{1620 \times 365 \times 24 \times 60 \times 60 \times 226} = 3.61 \times 10^{10}$

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A radioactive material decays by simultaneous emission of two particles with respective half lives 1620 and 810 years. The time (in years) after which one- fourth of the material remains is

1080

2430

3240

4860

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Explanation

(a) $λ = λ_{1} + λ_{2} \Rightarrow \frac{\ln 2}{T} = \frac{\ln 2}{T_{1}} + \frac{\ln 2}{T_{2}}$

$∴ T = \frac{T_{1} T_{2}}{T_{1} + T_{2}} = \frac{810 \times 1620}{810 + 1620} = 540 years$

Hence $\frac{1}{4} th$ of material remain after 1080 years.

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