Question

The binding energy per nucleon of ₃⁷Li and ₂⁴He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction ₃Li⁷ +₁H¹ → ₂He⁴ + ₂He⁴ + Q the value of energy Q released is:

पर्याय

• 8.4 MeV

• 17.3 MeV

• 19.6 MeV

• -2.4 MeV

Answer 1

17.3 MeV

Explanation:

Given, 375 Binding energy per nucleon of ₃Li⁷ an ₂He⁴ nuclei are 5.60 MeV and 7.06 MeV Energy released = 7.06 × 8 − 5.60 ×7

= 17.3 MeV