Nuclear Chemistry Question 13
Question: Consider an a-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
[UPSEAT 2001]
Options:
A) 23.8517× $ 10^{4}eV $
B) $ 26.147738\times 1{{0}^{4}}eV $
C) $ 25\text{.3522}\times 10^{4}eV $
D) $ 20.2254\times 10^{4}eV $
Show Answer
Answer:
Correct Answer: B
Solution:
$ {r _{nucleus}}=1\text{.3}\times 1{{0}^{\text{-13}}}\times {{(A)}^{1/3}}, $ where A is mass number $ {r _{U^{238}}}=1.3\times {{10}^{-13}}\times {{(238)}^{1/3}}=8.06\times {{10}^{-13}}cm. $
$ {r _{He^{4}}}=1.3\times {{10}^{-13}}\times {{(4)}^{1/3}}=2.06\times {{10}^{-13}}cm. $ \Total distance in between uranium and $ \alpha $ nuclei = 8. 06× $ {{10}^{-13}} $ + 2.06 × $ {{10}^{-13}} $ = 10.12 × $ {{10}^{-13}} $ cm Now repulsion energy = $ \frac{Q_1Q_2}{r}=\frac{92\times 4.8\times {{10}^{-10}}\times 2\times 4.8\times {{10}^{-10}}}{10.12\times {{10}^{-13}}}erg $
$ =418.9\times {{10}^{-7}}erg $ = $ 418.9\times {{10}^{-7}}\times 6.242\times 10^{11}eV $ = $ 26.147738\times 10^{4}eV. $
BETA
