A) \[11.35\times {{10}^{-21}}J\]
B) \[10.35\times {{10}^{-21}}J\]
C) \[15.35\times {{10}^{-21}}J\]
D) \[5.3\times {{10}^{-21}}J\]
Correct Answer: B
Solution :
Here, initial temperature \[{{T}_{1}}={{27}^{o}}C=300K\] Final temperature \[{{T}_{2}}={{227}^{o}}C=500K\]Average kinetic energy \[E=6.21\times {{10}^{-21}}J\] Average kinetic energy is given by \[E=\frac{3}{2}KT\propto T\] Hence \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{300}{500}=\frac{3}{5}\] \[{{E}_{2}}=\frac{5}{3}\times {{E}_{1}}=\frac{5}{3}\times 6.21\times {{10}^{-21}}\] \[{{E}_{2}}=10.35\times {{10}^{-21}}J\]
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