The distance of closest approach of an alpha | vedclass

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(C) At a sufficient distance,the electric potential energy is zero. So,$U_1 = 0$.At position $(1)$,the kinetic energy is $K_1 = \frac{1}{2} mv^2 = \fr

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$\frac{4 d}{9}$

Vedclass · Answer

(C) At a sufficient distance,the electric potential energy is zero. So,$U_1 = 0$.At position $(1)$,the kinetic energy is $K_1 = \frac{1}{2} mv^2 = \frac{P^2}{2m}$.At the distance of closest approach $(2)$,the kinetic energy is $K_2 = 0$.The electric potential energy at distance $d_c$ is $U_2 = \frac{1}{4 \pi \epsilon_0} \frac{q_\alpha q_n}{d_c}$.By the law of conservation of energy:$U_1 + K_1 = U_2 + K_2$$0 + \frac{P^2}{2m} = \frac{1}{4 \pi \epsilon_0} \frac{q_\alpha q_n}{d_c} + 0$From this,we see that the distance of closest approach $d_c \propto \frac{1}{P^2}$.Therefore,$\frac{d_2}{d_1} = \frac{P_1^2}{P_2^2}$.Given $d_1 = d$,$P_1 = P$,and $P_2 = 1.5 P = \frac{3}{2} P$:$d_2 = d 	imes \frac{P^2}{(1.5 P)^2} = d 	imes \frac{P^2}{2.25 P^2} = d 	imes \frac{1}{2.25} = d 	imes \frac{4}{9} = \frac{4d}{9}$.

The distance of closest approach of an alpha particle to a nucleus when the alpha particle moves towards the nucleus with linear momentum $P$ is $d$. What is the distance of closest approach of the alpha particle to the nucleus if the linear momentum of the alpha particle is $1.5 P$?

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