A) 2
B) \[\frac{1}{2}\]
C) 4
D) \[\frac{1}{4}\]
Correct Answer: B
Solution :
De-Broglie wavelength \[=\lambda =\frac{h}{\rho }\] \[\therefore \]\[\lambda =\frac{h}{\sqrt{2mk}}\Rightarrow \lambda \alpha \frac{1}{\sqrt{k}}\] Where, k is the kinetic energy of the electron. Now \[\lambda =\frac{h}{\sqrt{2mk}}\Rightarrow \lambda \alpha \frac{1}{\sqrt{k}}\] Hence, the correction option is [b].You need to login to perform this action.
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