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Question: A ray of energy 14.2 Me V is emitted from a <sup>60</sup>Co nucleus. The recoil energy of the Co nuc...

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

3 × 10-16J

Explanation

Solution

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

KE = p22m=E22c2mco\frac{p^{2}}{2m} = \frac{E^{2}}{2c^{2}m_{co}}

= (14×1.5×1013)22×(3×108×60×1.67×1027)\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