Modern Physics Ques 212
- The electrostatic energy of $Z$ protons uniformly distributed throughout a spherical nucleus of radius $R$ is given by
$E=\frac{3}{5} \frac{Z(Z-1) e^{2}}{4 \pi \varepsilon _0 R}$
The measured masses of the neutron, ${ } _1^{1} H,{ } _7^{15} N$ and ${ } _8^{15} O$ are $1.008665 $ $u, 1.007825 $ $u, 15.000109 $ $u$ and $15.003065 $ $u$, respectively. Given that the radii of both the ${ } _7^{15} N$ and ${ } _8^{15} O$ nuclei are same, $1 u=931.5 $ $MeV / c^{2}$ ( $c$ is the speed of light) and $e^{2} /\left(4 \pi \varepsilon _0\right)=1.44$ $ MeV fm$. Assuming that the difference between the binding energies of ${ } _7^{15} N$ and ${ } _8^{15} O$ is purely due to the electrostatic energy, the radius of either of the nuclei is ( $\left.1 fm=10^{-15} m\right)$
(2016 Adv.)
(a) $2.85 $ $fm$
(b) $3.03 $ $fm$
(c) $3.42 $ $fm$
(d) $3.80 $ $fm$
Show Answer
Answer:
Correct Answer: 212.(c)
Solution:
Formula:
- Electrostatic energy
$=$ Binding energy of $N-$ Binding energy of $O$
$=\left[\left[7 M _H+8 M _n-M _N\right]-\left[8 M _H+7 M _n-M _O\right]\right] \times C^{2}$
$=\left[-M _H+M _n+M _O M _N\right] C^{2}$
$=[-1.007825+1.008665+15.003065-15.000109] \times 93.15$
$=+3.5359 $ $MeV$
$ \begin{aligned} \Delta E & =\frac{3}{5} \times \frac{1.44 \times 8 \times 7}{R}=3.5359 \\ R & =\frac{3 \times 1.44 \times 14}{5 \times 3.5359}=3.42 fm \end{aligned} $
BETA
