Question

# A proton when accelerated through a potential difference of $$V$$, has a de Broglie wavelength $$\lambda$$ associated with it. If an alpha particle is to have the same de Broglie wavelength $$\lambda$$, it must be accelerated through a potential difference of:

A
V8
B
V4
C
4V
D
8V

Solution

## The correct option is B $$\displaystyle\frac{V}{8}$$Since the kinetic energy will be equal to the work done on particle,$$K=qV$$Also, $$p=\sqrt { 2mK }$$Thus, $$\lambda =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mK } } =\dfrac { h }{ \sqrt { 2mqV } }$$For both to have same $$\lambda$$,$${ m }_{ p }{ q }_{ p }{ V }={ m }_{ a }{ q }_{ a }{ V }_{ a }$$$${ V }_{ a }=\dfrac { 1 }{ (4)(2) } V=\dfrac { V }{ 8 }$$Answer is option A.Physics

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