Question

An electron accelerated through a potential of \(10,000 \mathrm{~V}\) from rest has a de-Broglie wave length ' \(\lambda\) '. What should be the accelerating potential so that the wave length is doubled?
(A) \(20,000 \mathrm{~V}\)
(B) \(40,000 \mathrm{~V}\)
(C) \(5,000 \mathrm{~V}\)
(D) \(2,500 \mathrm{~V}\)

Answer 1

Ans: (D)
 As \(\lambda \propto \frac{1}{\sqrt{V}}\)
$$
\frac{V^{\prime}}{V}=\frac{1}{4}, \text { or } V^{\prime}=\frac{10000}{4}=2500 \mathrm{~V}
$$

Answer 2

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