A) \[{{E}^{2}}\]
B) \[{{E}^{1/2}}\]
C) \[{{E}^{-1}}\]
D) \[{{E}^{-1/2}}\]
Correct Answer: D
Solution :
[d] For photon, \[{{\lambda }_{2}}=\frac{hc}{E}\] ?.(i) For proton, \[p=\sqrt{2mE}\] \[{{\lambda }_{1}}=\frac{h}{p}=\frac{h}{\sqrt{2mE}}\] ?(ii) \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{hc}{E\times \frac{h}{\sqrt{2mE}}}\propto {{E}^{-1/2}}\]You need to login to perform this action.
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