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(B) The radius of a nucleus is given by $r = 1.3 	imes 10^{-13} 	imes (A)^{1/3} \ cm$,where $A$ is the mass number.For $U^{238}$,$r_{U} = 1.3 	imes

Vedclass · Accepted Answer

$26.147738 	imes 10^4 \ eV$

Vedclass · Answer

(B) The radius of a nucleus is given by $r = 1.3 	imes 10^{-13} 	imes (A)^{1/3} \ cm$,where $A$ is the mass number.For $U^{238}$,$r_{U} = 1.3 	imes 10^{-13} 	imes (238)^{1/3} \approx 8.06 	imes 10^{-13} \ cm$.For the $\alpha$-particle $(He^{4})$,$r_{\alpha} = 1.3 	imes 10^{-13} 	imes (4)^{1/3} \approx 2.06 	imes 10^{-13} \ cm$.The total distance $r$ between the centers of the nuclei is $r = r_{U} + r_{\alpha} = 8.06 	imes 10^{-13} + 2.06 	imes 10^{-13} = 10.12 	imes 10^{-13} \ cm$.The coulombic repulsion energy is $E = \frac{k q_1 q_2}{r}$ (in $CGS$ units,$k=1$).$E = \frac{(Z_1 e)(Z_2 e)}{r} = \frac{(92 	imes 4.8 	imes 10^{-10} \ esu) 	imes (2 	imes 4.8 	imes 10^{-10} \ esu)}{10.12 	imes 10^{-13} \ cm} \ erg$.$E = \frac{423.936 	imes 10^{-20}}{10.12 	imes 10^{-13}} \ erg \approx 4.189 	imes 10^{-5} \ erg$.Converting to $eV$ $(1 \ erg = 6.242 	imes 10^{11} \ eV)$:$E = 4.189 	imes 10^{-5} 	imes 6.242 	imes 10^{11} \ eV \approx 26.147738 	imes 10^4 \ eV$.

Consider an $\alpha$-particle just in contact with a $_{92}U^{238}$ nucleus. Calculate the coulombic repulsion energy (i.e.,the height of the coulombic barrier between $U^{238}$ and $\alpha$-particle) assuming that the distance between them is equal to the sum of their radii.

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