A) \[{{E}^{o}}\]
B) \[{{E}^{1/2}}\]
C) \[{{E}^{-1}}\]
D) \[{{E}^{-2}}\]
Correct Answer: B
Solution :
\[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=[h/{{(2mE)}^{1/2}}]/[hc/E]\] Thus, \[{{\lambda }_{1}}/{{\lambda }_{2}}\propto {{E}^{1/2}}\]You need to login to perform this action.
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