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Question: In the reaction <sub>1</sub>H<sup>2</sup> + <sub>1</sub>H<sup>3</sup>¾®<sub>2</sub>He<sup>4</sup> +<...

In the reaction 1H2 + 1H3¾®2He4 +0n1 if the binding energies of 1H2, 1H3 and 2He4 are respectively a, b and c (in MeV), then the energy(in MeV) released in this reaction is-

A

a + b + c

B

a + b – c

C

c – a – b

D

c + a – b

Answer

c – a – b

Explanation

Solution

1H2 + 1H3 ¾® 2He4 + 0n1

a b c

DE = c – (a + b)

DE = c – a – b