A) \[{{10}^{7}}K\]
B) \[{{10}^{5}}K\]
C) \[{{10}^{3}}K\]
D) \[{{10}^{9}}K\]
Correct Answer: D
Solution :
[d] average kinetic energy per molecule \[=\frac{3}{2}kT\] This kinetic energy should be able to provide the repulsive potential energy \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{2}kT=7.7\times {{10}^{-14}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,T=\frac{2\times 7.7\times {{10}^{-14}}}{3\times 1.38\times {{10}^{-23}}}=3.7\times {{10}^{9}}K\]You need to login to perform this action.
You will be redirected in
3 sec