If $λ_0$ is the de Broglie wavelength for a proton accelerated through a potential difference of 100 V, the de Broglie wavelength for α−particle accelerated through the same potential difference is:

$\frac{λ_0}{2\sqrt{2}}$

Kinetic energy gained by a charge q after being accelerated through a potential difference V is given by

$qV=\frac{1}{2}mv^2;v=\sqrt{\frac{2qV}{m}}$ and $mv=\sqrt{2mqV}$

de Broglie wavelength, $λ=\frac{h}{mv}=\frac{h}{\sqrt{2mqV}}$

For a proton, $λ_p=\frac{h}{\sqrt{2m_pq_pV_p}}$

For an α − particle

$λ_α=\frac{h}{\sqrt{2m_αq_αV_α}}∴\frac{λ_α}{λ_p}=\sqrt{\frac{2m_pq_pV_p}{2m_αq_αV_α}}$

$V_α=V_p=100V$ (Given)

$∴\frac{λ_α}{λ_p}=\sqrt{\frac{m_pq_p}{m_αq_α}}=\sqrt{\frac{m_pq_p}{(4m_p)(2q_p)}}=\sqrt{\frac{1}{8}}=\frac{1}{2\sqrt{2}}$

$λ_α=\frac{λ_p}{2\sqrt{2}}=\frac{λ_0}{2\sqrt{2}}$ $(∵λ_p=λ_0(Given))$