Question

The energy released in joule and MeV in the following nuclear reaction,

\[\ce{^2_1{H} + ^2_1{H} -> ^3_2{He} + ^1_0{n}}\]

(Assume that the masses of \[\ce{^2_1{H}}\], \[\ce{^3_2{He}}\] and neutron (n) respectively are 2.0141, 3.0160 and 1.0087 in amu) is ______ × 10⁻¹³ J.

Options

• 1.87

• 2.65

• 3.24

• 5.22

Answer 1

The energy released in joule and MeV in the following nuclear reaction,

\[\ce{^2_1{H} + ^2_1{H} -> ^3_2{He} + ^1_0{n}}\]

(Assume that the masses of \[\ce{^2_1{H}}\], \[\ce{^3_2{He}}\] and neutron (n) respectively are 2.0141, 3.0160 and 1.0087 in amu) is 5.22 × 10⁻¹³ J.

Explanation:

Δm = [2 × 2.0141] – 3.0160 – 1.0087

= 3.5 × 10^–3 amu

ΔE = Δm × 931.478 MeV

ΔE = 3.5 × 10^–3 × 931.478

= 3.260 MeV

ΔE = 5.22 × 10^–13 J