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A gamma ray photon creates an electron- positron pair. If the rest mass energy of an electron is 0.5 MeV and the total K.E. of the electron- positron pair is 0.78 MeV, then the energy of the gamma ray photon must be

0.78 MeV

1.78 MeV

1.28 MeV

0.28 MeV

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Explanation

(b) Energy of $γ$ -ray photon = 0.5+0.5+0.78 = 1.78 MeV

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Which of the following statement is true

${}_{78}{Pt}_{192}$ has 78 neutrons

${}_{84}{Po}_{214} \to {}_{82}{Pb}_{210} + β^{-}$

${}_{92}U_{238} \to {}_{90}{Th}_{234} + {}_{2}{He}_{4}$

${}_{90}{Th}_{234} \to {}_{91}{Pa}_{234} + {}_{2}{He}_{4}$

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Explanation

(c) (a) ${}_{78}{Pt}^{192}$ has 114 neutrons
(b) ${}_{84}{Po}^{214} \to {}_{83}{Pb}^{210} + β^{-}$
(c) ${}_{9}U^{238} \to {}_{90}{Th}^{234} + {}_{2}{He}^{4}$
(d) ${}_{90}{Th}^{234} \to {}_{91}{Pa}^{234} + {}_{2}{He}^{4}$

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Fast neutrons can easily be slowed down by

The use of lead shielding

Passing them through water

Elastic collisions with heavy nuclei

Applying a strong electric field

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Explanation

(b) Heavy water is used as moderators in nuclear reactions to slow down the neutrons.

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When ${}_{92}U_{235}$ undergoes fission, 0.1% of its original mass is changed into energy. How much energy is released if 1kg of ${}_{92}U_{235}$ undergoes fission

$9 \times 10^{10} J$

$9 \times 10^{11} J$

$9 \times 10^{12} J$

$9 \times 10^{13} J$

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Explanation

(d) $E = ∆ {mc}^{2}, ∆ m = \frac{0.1}{100} = 10^{- 3} kg$

$∴ E = 10^{- 3} \times {(3 \times 10^{8})}^{2} = 10^{- 3} \times 9 \times 10^{16} = 9 \times 10^{13} J$

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$γ$ -rays radiation can be used to create electron-positron pair. In this process of pair production, $γ$ -rays energy cannot be less than

5.0 MeV

4.02 MeV

15.0 MeV

1.02 MeV

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Explanation

(d) Energy released by $γ$ -rays for pair production must be greater than 1.02 MeV.

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In the nuclear reaction ${}_{92}U_{238} \to {}_{z}{Th}_{A} + {}_{2}{He}_{4}$ , the values of A and Z are

A = 234, Z = 94

A = 234, Z = 90

A = 238, Z = 94

A = 238, Z = 90

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Explanation

(b) A = 238 – 4 = 234 and Z = 92 – 2 = 90.

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If 200 MeV energy is released in the fission of a single $U^{235}$ nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given $1 eV = 1.6 \times 10^{- 19} J$ )

$3.125 \times 10^{13}$

$3.125 \times 10^{14}$

$3.125 \times 10^{15}$

$3.125 \times 10^{16}$

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Explanation

(a) $P = n (\frac{E}{t}) \Rightarrow 1000 = \frac{n \times 200 \times 10^{6} \times 1.6 \times 10^{- 19}}{t} \Rightarrow \frac{n}{t} = 3.125 \times 10^{13}$

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In nuclear fission, the fission reactions proceeds with a projectile. Which of the following suits the best

Slow proton

Fast neutron

Slow neutron

None of these

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Explanation

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In the nuclear reaction ${}_{85}X_{297} \to Y + 4 α$ , Y is

${}_{76}Y_{287}$

${}_{77}Y_{285}$

${}_{77}Y_{281}$

${}_{77}Y_{289}$

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Explanation

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Thermal neutrons can cause fission in

$U^{235}$

$U^{238}$

${Pu}^{238}$

${Th}^{232}$

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Explanation

(a) For nuclei containing an even no. of electrons, fission can only occur if the incident neutrons have energy above about 1 MeV.

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