A) \[\frac{1}{c}{{\left( \frac{E}{2m} \right)}^{\frac{1}{2}}}\]
B) \[{{\left( \frac{E}{2m} \right)}^{\frac{1}{2}}}\]
C) \[c{{\left( 2mE \right)}^{\frac{1}{2}}}\]
D) \[\frac{1}{xc}{{\left( \frac{E}{2m} \right)}^{\frac{1}{2}}}\]
Correct Answer: A
Solution :
[a] For electron De-Broglie wavelength, \[{{\lambda }_{e}}=\frac{h}{\sqrt{2mE}}\]; For photon E = pc \[\Rightarrow \]De-Broglie wavelength, \[{{\lambda }_{Ph}}=\frac{hc}{E}\] \[\therefore \frac{{{\lambda }_{e}}}{{{\lambda }_{Ph}}}=\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}={{\left( \frac{E}{2m} \right)}^{1/2}}\frac{1}{c}\]
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