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Energy generation in stars is mainly due to

Chemical reactions

Fission of heavy nuclei

Fusion of light nuclei

Fusion of heavy nuclei

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Explanation

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Fusion reaction is initiated with the help of

Low temperature

High temperature

Neutrons

Any particle

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Explanation

(b) Fusion reaction requires a very high temperature =(10⁷ K).

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When a ${}_{4}{Be}_{9}$ atom is bombarded with $α$ particles, one of the product of nuclear transmutation is ${}_{6}C_{12}$ . The other is

${}_{- 1}e_{0}$

${}_{1}H_{1}$

${}_{1}D_{2}$

${}_{0}n_{1}$

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Explanation

(d) ${}_{4}{Be}_{9} + {}_{2}{He}_{4} \to {}_{6}C_{12} + {}_{0}n_{1}$

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During the nuclear fusion reaction

A heavy nucleus breaks into two fragments by itself

A light nucleus bombarded by thermal neutrons breaks up

A heavy nucleus bombarded by thermal neutrons breaks up

Two light nuclei combine to give a heavier nucleus and
possibly other products

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Explanation

(d) During the nuclear fusion reaction two light nuclei combine to give a heavy nuclei and possible other products.

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The main source of solar energy is

Fission reactions

Fusion reactions

Chemical reactions

Combustion reactions

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Explanation

(b) The main source of solar energy is fusion reactions of hydrogen atoms.

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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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