A) 1 : 2
B) 2\[\sqrt{2}\]: 1
C) 2 : 1
D) \[\sqrt{2}\]: 1
Correct Answer: B
Solution :
Here, \[qV=\frac{1}{2}m{{v}^{2}}\] or \[mv=\sqrt{2qmV}\] So, de-Broglie wavelength, \[\lambda =\frac{h}{mv}=\frac{h}{\sqrt{2qmV}}\]ie, \[\lambda \propto \frac{1}{\sqrt{qm}}\] Hence, \[\frac{{{\lambda }_{p}}}{{{\lambda }_{a}}}=\sqrt{\frac{2e\times 4m}{e\times m}}=2\sqrt{2}\]You need to login to perform this action.
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