A) 23.8517×\[{{10}^{4}}eV\]
B) \[26.147738\times \text{1}{{\text{0}}^{\text{4}}}eV\]
C) \[\text{25}\text{.3522}\times {{10}^{4}}eV\]
D) \[20.2254\times {{10}^{4}}eV\]
Correct Answer: B
Solution :
\[{{r}_{\text{nucleus}}}=\text{1}\text{.3}\times \text{1}{{\text{0}}^{\text{-13}}}\times {{(A)}^{1/3}},\] where A is mass number \[{{r}_{{{U}^{238}}}}=1.3\times {{10}^{-13}}\times {{(238)}^{1/3}}=8.06\times {{10}^{-13}}cm.\] \[{{r}_{H{{e}^{4}}}}=1.3\times {{10}^{-13}}\times {{(4)}^{1/3}}=2.06\times {{10}^{-13}}cm.\] \Total distance in between uranium and \[\alpha \] nuclei = 8. 06× \[{{10}^{-13}}\] + 2.06 × \[{{10}^{-13}}\] = 10.12 × \[{{10}^{-13}}\]cm Now repulsion energy = \[\frac{{{Q}_{1}}{{Q}_{2}}}{r}=\frac{92\times 4.8\times {{10}^{-10}}\times 2\times 4.8\times {{10}^{-10}}}{10.12\times {{10}^{-13}}}erg\] \[=418.9\times {{10}^{-7}}erg\]= \[418.9\times {{10}^{-7}}\times 6.242\times {{10}^{11}}eV\] = \[26.147738\times {{10}^{4}}eV.\]
You need to login to perform this action.
You will be redirected in
3 sec
