A free nucleus of mass $24 a m u$ , intially at rest emits a gamma photon. The energy of the photon is $7 M e V$ . The kinetic energy of recoiling nucleus is approximately equal to [Take; $1 a m u = 9.31 M e V$ ]

3.3 k e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.4 k e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.1 k e V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.5 k e V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1.1 k e V$
$E = \frac{P^{2}}{2 m}$

conservation of momentum for photon :
$E = \frac{h c}{λ} = 7 M e V$
$p = \frac{h}{λ} = 7 / c M e V$

Equating the momentum:
$\frac{7}{c} = \sqrt{2 E_{n u c l e u s} m}$

Substitute = 24 amu,
Solving with appropriate units:
$E_{n u c l e u s} = 1.1 K e V$

Suggest Corrections

Similar questions

24 a m u

7 M e V

k e V

24 a m u

7 M e V

20 u

γ

6 M e V

1 u = 1.6 \times 10^{- 27} k g

M

v

Join BYJU'S Learning Program