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Nuclei MCQs for NEET — Physics Questions with Answers

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Mean life of a radioactive sample is 100 seconds. Then its half life (in minutes) is

0.693

$10^{- 4}$

1.155

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Explanation

(d) Mean life (T) = 1/ $λ$ = 100 second
Half-life = $\frac{0.693}{λ} = \frac{0.693 \times 100}{60} = 1.155 \min$

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${}_{86}^{222}A \to {}_{84}^{210}B$ . In this reaction how many $α$ and $β$ particles are emitted

6 $α$ , 3 $β$

3 $α$ , 4 $β$

4 $α$ , 3 $β$

3 $α$ , 6 $β$

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Explanation

(b) By using $n_{α} = \frac{A - A'}{4}$ and $n_{β} = 2 n_{α} - Z + Z'$

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The phenomenon of radioactivity is

Exothermic change which increases or decreases with temperature

Increases on applied pressure

Nuclear process does not depend on external factors

None of the above

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Explanation

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If half life of radium is 77 days. Its decay constant in day will be

$3 \times 10^{- 13}$ /day

$9 \times 10^{- 3}$ /day

$1 \times 10^{- 3}$ /day

$6 \times 10^{- 3}$ /day

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Explanation

(b) $λ = \frac{0.693}{T_{1 / 2}} = \frac{0.693}{77} = 9 \times 10^{- 3} / day$

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In a sample of radioactive material, what fraction of the initial number of active nuclei will remain undisintegrated after half of a half-life of the sample

$\frac{1}{4}$

$\frac{1}{2 \sqrt{2}}$

$\frac{1}{\sqrt{2}}$

$2 \sqrt{2}$

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Explanation

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Consider two nuclei of the same radioactive nuclide. One of the nuclei was created in a supernova explosion 5 billion years ago. The other was created in a nuclear reactor 5 minutes ago. The probability of decay during the next time is

Different for each nuclei

Nuclei created in explosion decays first

Nuclei created in the reactor decays first

Independent of the time of creation

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Explanation

(d) The half life and decay constant, independent of time of creation of radioactive nuclei.

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A neutron with velocity V strikes a stationary deuterium atom. Its kinetic energy changes by a factor of

$\frac{15}{16}$

$\frac{1}{2}$

$\frac{2}{1}$

None of these

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The sun radiates energy in all directions. The average radiations received on the earth surface from the sun is 1.4 $kilowatt / m^{2}$ .The average earth- sun distance is $1.5 \times 10^{11}$ metres. The mass lost by the sun per day is
(1 day = 86400 seconds)

$4.4 \times 10^{9} kg$

$7.6 \times 10^{14} kg$

$3.8 \times 10^{12} kg$

$3.8 \times 10^{14} kg$

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Explanation

(d) Energy radiated = 1.4 $kW / m^{2}$

$= 1.4 kJ / \sec m^{2} = \frac{1.4 kJ}{\frac{1}{86400} day m^{2}} = \frac{1.4 \times 86400}{day m^{2}}$

Total energy radiated/day

$= \frac{4 π \times {(1.5 \times 10^{11})}^{2} \times 1.4 \times 86400}{1} \frac{kJ}{day} = E$

$∴ E = {mc}^{2} \Rightarrow m = \frac{E}{c^{2}}$

$= \frac{4 π \times {(1.5 \times 10^{11})}^{2} \times 1.4 \times 86400}{{(3 \times 10^{8})}^{2}} = 3.8 \times 10^{14} kg$

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The binding energy per nucleon of $O^{16}$ is 7.97 MeV and that of $O^{17}$ is 7.75 MeV. The energy (in MeV) required to remove a neutron from $O^{17}$ is

3.52

3.64

4.23

7.86

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Explanation

(c) The equation is $O^{17} \to {}_{0}^{1}n + O^{16}$
$∴$ Energy required = B.E. of $O^{17}$ – B.E. of $O^{16}$
= 17 $\times$ 7.75 – 16 $\times$ 7.97 = 4.23 MeV

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The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity 0.5 c. The change in its energy will be

0.026 MeV

0.051 MeV

0.079 MeV

0.105 MeV

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Explanation

$= m_{0} c^{2} (\frac{1}{\sqrt{1 - (v^{2} / c^{2})}} - 1) = 0.511 (\frac{1}{\sqrt{0.75}} - 1)$

= 0.079 MeV

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