Question

A proton and α-particle are accelerated through different potentials V₁ and V₂ respectively so that they have the same de Broglie wavelengths. Find ${\displaystyle \frac{V_1}{V_2}}$.

Answer 1

De-Broglie wavelength is given as

${\displaystyle \lambda =\frac{h}{p}=\frac{h}{\sqrt{2mqV}}}$

Where m is the mass of particle,

q is the charge of particle

and V is the accelerating voltage.

De-Broglie wavelength of proton and α-particle is the same but different accelerating voltage.

⇒ λ_P = λ_α 

⇒ ${\displaystyle \frac{h}{\sqrt{2m_p{q}_p{V}_1}}=\frac{h}{\sqrt{2m_{\alpha }{q}_{\alpha }{V}_2}}}$

⇒ ${\displaystyle \frac{V_1}{V_2}=\frac{m_{\alpha }{q}_{\alpha }}{m_p{q}_p}=\frac{4m_p\cdot 2e}{m_p\cdot e}=8 ...\left[\begin{array}{c}\because m_{\alpha }=4m_p\\ q_{\alpha }=22q_p\end{array}\right]}$

Hence, ${\displaystyle \frac{V_1}{V_2}=\frac{8}{1}}$