JEE Main 2019 · Physics · Atoms, Nuclei and Dual Nature of Radiation · easy

In the reaction ^21H+^21H→^32He+^10n, the atomic masses are 2.0141 u for deuterium, 3.0160 u for helium-3 and 1.0087 u for the neutron (1 u=931.5…

Single correct option

In the reaction

{}^2_1\mathrm{H}+{}^2_1\mathrm{H}\to{}^3_2\mathrm{He}+{}^1_0n

, the atomic masses are

2.0141\ \mathrm{u}

for deuterium,

3.0160\ \mathrm{u}

for helium-3 and

1.0087\ \mathrm{u}

for the neutron (

1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^2

). The reacting deuterons are nearly at rest. The kinetic energy carried away by the neutron is nearly

Show answer and solution

Answer: (C) $2.44\ \mathrm{MeV}$

The mass defect is

2(2.0141)-(3.0160+1.0087)=0.0035\ \mathrm{u}

, so

Q=0.0035\times931.5\approx3.26\ \mathrm{MeV}

. With initial momentum zero, the products have equal and opposite momenta

p

, and

K=p^2/2m

means the kinetic energies are inversely proportional to the masses. The neutron gets the fraction

m_{He}/(m_{He}+m_n)=3.016/4.0247

Q

K_n\approx3.26\times0.749\approx2.44\ \mathrm{MeV}

. The value

3.26\ \mathrm{MeV}

is the total released energy, not the neutron's share.

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake