Question

A proton when accelerated through a potential difference of V volt has wavelength $\lambda$ associated with it .An electron to have the same $\lambda$ must be accelerated through a p.d of

• A. $\frac{V}{8}$ volt
• B. 4V volt
• C. 2V volt
• D. 1838V volt

Answer 1

The correct option is D 1838V volt

The energy gained by the proton will be

$E=q_{o}V=\frac{p^2}{2m_p}$ ...... ( 1 )

where $q_o$ is the charge on the proton, V is the p.d applied, p is the momentum of the proton and $m_p$ is the mass of the proton.

The momentum p of the proton is given by de Broglie's equation as

$p=\frac{h}{\lambda }$

Therefore, for an electron to have the same $\lambda$, it must have the same momentum.

$\therefore$ $E_e=\frac{p^2}{2m_e}$

Using p from equation 1,

$⟹$ $E_e=\frac{m_p{q}_{o}V}{m_e}$

$⟹$ $E_e=q_{o}V\left(\frac{m_p}{m_e}\right)=q_o\left(1838V\right)$

Therefore, an electron must be accelerated through a p.d of 1838V volt in order to achieve a wavelength $\lambda$.