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. $\frac{V}{8}$
• B. $\frac{V}{4}$
• C. $4V$
• D. $8V$

Answer 1

Answer: (A)

$\frac{V}{8}$

Since the kinetic energy will be equal to the work done on particle,

$K=qV$

Also, $p=\sqrt{2mK}$

Thus, $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mqV}}$

For both to have same $\lambda$,

$m_p{q}_{p}V=m_a{q}_a{V}_a$

$V_a=\frac{1}{\left(4\right)\left(2\right)}V=\frac{V}{8}$

Answer is option A.