Question

A ray of energy 14.2 Me V is emitted from a ⁶⁰Co nucleus. The recoil energy of the Co nucleus is nearly

• A. 3 × 10^-16J
• B. 3 × 10^-15J
• C. 3 × 10^-14J
• D. 3 × 10^-13J

Answer 1

Answer: (A)

3 × 10^-16J

Answer 2

Pr = Pco' Pr = $\frac{E}{c},$

KE = $\frac{p^{2}}{2m} = \frac{E^{2}}{2c^{2}m_{co}}$

= $\frac{\left( 14 \times 1.5 \times 10^{- 13} \right)^{2}}{2 \times \left( 3 \times 10^{8} \times 60 \times 1.67 \times 10^{- 27} \right)}$ = 3 × 10^-16J