A) \[0.186\times {{10}^{-20}}\]Joule
B) \[0.372\times {{10}^{-20}}\]Joule
C) \[0.56\times {{10}^{-20}}\] Joule
D) \[5.6\times {{10}^{-20}}\]Joule
Correct Answer: C
Solution :
[c] \[{{E}_{av}}=\frac{f}{2}kT=\frac{3}{2}\times 1.38\times {{10}^{-23}}\times 273\]\[=0.56\times {{10}^{-20}}J\]You need to login to perform this action.
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