The nucleus $^{235}_{92}\mathrm{Y}$, initially at rest, decays into $^{231}_{90}\mathrm{X}$ by $\alpha$-emission: $^{235}_{92}\mathrm{Y} \rightarrow\ ^{231}_{90}\mathrm{X} +\ ^{4}_{2}\mathrm{He} + \text{energy}$. The binding energies per nucleon of the parent nucleus, the daughter nucleus and the $\alpha$-particle are 7.8 MeV, 7.835 MeV and 7.07 MeV respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted $\alpha$-particle. (Mass of $\alpha$-particle $= 6.68\times10^{-27}\ \mathrm{kg}$.)

$Q = 5.165\ \text{MeV}$; $v_\alpha \approx 1.57\times10^{7}\ \text{m/s}$.

CBSE Class 12 · 2020 · Modern PhysicsShort Answer

The nucleus

_{92}^{235} Y

, initially at rest, decays into

_{90}^{231} X

α

-emission:

_{92}^{235} Y \to_{90}^{231} X +_{2}^{4} He + energy

. The binding energies per nucleon of the parent nucleus, the daughter nucleus and the

α

-particle are 7.8 MeV, 7.835 MeV and 7.07 MeV respectively. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share in the energy of the reaction, find the speed of the emitted

α

-particle. (Mass of

α

-particle

= 6.68 \times 1 0^{- 27} kg

Answer: $Q = 5.165 MeV$ ; $v_{α} \approx 1.57 \times 1 0^{7} m/s$ .

Solution

Energy released

=

(total BE of products)

-

(total BE of parent):

Q = [231 (7.835) + 4 (7.07)] - 235 (7.8) = (1809.885 + 28.28) - 1833 = 5.165 MeV .

Neglecting the recoil of the daughter, this energy is carried off as the kinetic energy of the

α

-particle:

K_{α} = 5.165 MeV = 5.165 \times 1.6 \times 1 0^{- 13} = 8.26 \times 1 0^{- 13} J .

v = \frac{2 K _{α}}{m _{α}} = \frac{2 ( 8.26 \times 1 0 ^{- 13} )}{6.68 \times 1 0 ^{- 27}} \approx 1.57 \times 1 0^{7} m/s .

v \approx 1.57 \times 1 0^{7} m/s

CBSE Class 12 Board · 2020 · Subjective

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CBSE Class 12 · 2020 · Modern PhysicsShort Answer

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