Solveeit Logo
Login

Question

Question: In the reaction \({}_{1}^{2}H\) + \({}_{1}^{3}H\) ®\({}_{2}^{4}He\) + \({}_{0}^{1}n\). If the bindin...

In the reaction 12H{}_{1}^{2}H + 13H{}_{1}^{3}H ®24He{}_{2}^{4}He + 01n{}_{0}^{1}n. If the binding energies of 12H{}_{1}^{2}H, 13H{}_{1}^{3}H and 24He{}_{2}^{4}He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is –

A

c + a – b

B

c – a – b

C

a + b + c

D

a + b – c

Answer

c – a – b

Explanation

Solution

E = B.E.Product – B.E.Reactant

= (B.E.2He4 + B.E.0n1) – (B.E.1H2 + B.E.1H3)

= (c + 0) – (a + b)

= c – (a + b)

E=c−a−b\mathrm { E } = \mathrm { c } - \mathrm { a } - \mathrm { b }