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Physics MCQs for NEET — Practice Questions with Answers

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If an electron and a positron annihiliate, then the energy released is

$3.2 \times 10^{- 13} J$

$1.6 \times 10^{- 13} J$

$4.8 \times 10^{- 13} J$

$6.4 \times 10^{- 13} J$

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Explanation

(b) Mass of electron = mass of positron = $9.1 \times 10^{- 31}$ kg
Energy released $E = (2 m) c^{2}$
= $2 \times 9.1 \times 10^{- 31} \times (3 \times 10^{8})^{2} = 1.6 \times 10^{- 13} J$

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The energy liberated on complete fission of 1 kg of ${}_{92}^{235}U$ is (Assume 200 MeV energy is liberated on fission of 1 nucleus)

$8.2 \times 10^{10} J$

$8.2 \times 10^{9} J$

$8.2 \times 10^{13} J$

$8.2 \times 10^{16} J$

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Explanation

(c) Mass of a uranium nucleus $= 92 \times 1.6725 \times 10^{- 27} + 143 \times 1.6747 \times 10^{- 27} = 393.35 \times 10^{- 27} kg$

Number of nuclei in the given mass $= \frac{1}{393.35 \times 10^{- 27}} = 2.542 \times 10^{24}$

Energy released $= 200 \times 2.542 \times 10^{24} MeV$

$= 5.08 \times 10^{26} MeV = 8.135 \times 10^{13} J = 8.2 \times 10^{13} J$

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The nuclear reaction ${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{4}{He}$ (mass of deuteron = 2.0141 a.m.u. and mass of He = 4.0024 a.m.u.) is

Fusion reaction releasing 24 MeV energy

Fusion reaction absorbing 24 MeV energy

Fission reaction releasing 0.0258 MeV energy

Fission reaction absorbing 0.0258 MeV energy

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Explanation

(a) Total mass of reactants $= (2.0141) \times 2 = 4.0282 amu$
Total mass of products =4.0024 amu
Mass defect = 4.0282 amu-4.0024 amu = 0.0258 amu

$∴$ Energy released E = 931 $\times$ 0.0258 = 24 MeV

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In the following reaction the value of ‘X’ is
${}_{7}^{14}N + {}_{2}^{4}{He} \to X + {}_{1}^{1}H$

${}_{8}^{17}N$

${}_{8}^{17}O$

${}_{7}^{16}O$

${}_{7}^{16}N$

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Explanation

(b) ${}_{7}^{14}N + {}_{2}^{4}{He} \to {}_{8}^{17}O + {}_{1}^{1}H$

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Which of the following are suitable for the fusion process

Heavy nuclei

Light nuclei

Atom bomb

Radioactive decay

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Explanation

(b) Fusion process results in increase in mass number. For light nuclei, fusion process increases the binding energy per nucleon which increases stability. So light nuclei are suitable for fusion process.

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A deutron is bombarded on ${}_{8}^{16}O$ nucleus and $α$ -particle is emitted. The product nucleus is

${}_{7}^{13}N$

${}_{5}^{10}B$

${}_{4}^{9}{Be}$

${}_{7}^{14}N$

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Explanation

(d) ${}_{8}^{16}O + {}_{1}^{2}H \to {}_{7}^{14}N + {}_{2}^{4}{He}$

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A nuclear reaction given by
${}_{Z}^{A}X \to {}_{Z + 1}^{A}Y + {}_{- 1}^{0}e + \bar{p}$ represents

$γ$ -decay

Fusion

Fission

$β$ -decay

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Explanation

(d) It is a $β$ -decay

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The binding energy of nucleus is a measure of its

Charge

Mass

Momentum

Stability

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Explanation

(d) B.E. per nucleon ∝ stability.

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If M is the atomic mass and A is the mass number, packing fraction is given by

$\frac{A}{M - A}$

$\frac{A - M}{A}$

$\frac{M}{M - A}$

$\frac{M - A}{A}$

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Explanation

(d) Packing fraction = $\frac{M - A}{A}$

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$M_{p}$ denotes the mass of a proton and $M_{n}$ that of a neutron. A given nucleus, of binding energy B, contains Z protons and N neutrons. The mass M(N, Z) of the nucleus is given by (c is the velocity of light)

$M (N, Z) = {NM}_{n} + {ZM}_{p} - {Bc}^{2}$

$M (N, Z) = {NM}_{n} + {ZM}_{p} + {Bc}^{2}$

$M (N, Z) = {NM}_{n} + {ZM}_{p} - B / c^{2}$

$M (N, Z) = {NM}_{n} + {ZM}_{p} + B / c^{2}$

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Explanation

$\Rightarrow M (N, Z) = {ZM}_{p} + {NM}_{n} - B / c^{2}$

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