A) \[\lambda =\frac{h}{\sqrt{mK}}\]
B) \[\lambda =\frac{2h}{\sqrt{mK}}\]
C) \[\lambda =\frac{h}{2\sqrt{mK}}\]
D) \[\lambda =\frac{h}{\sqrt{2\,mK}}\]
Correct Answer: D
Solution :
The energy of a charged particle accelerated through potential difference V is \[K=\frac{1}{2}m{{v}^{2}}=qV\] Hence, de-Broglie wavelength \[\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}\]
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