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 `V_1/V_2`.

Answer 1

De-Broglie wavelength is given as

`lambda = h/p = 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 

⇒ `h/(sqrt(2m_pq_pV_1)) = h/(sqrt(2m_alphaq_alphaV_2))`

⇒ `V_1/V_2 = (m_alphaq_alpha)/(m_pq_p) = (4m_p * 2e)/(m_p * e) = 8    ...[(becausem_alpha = 4m_p),(q_alpha = 22q_p)]`

Hence, `V_1/V_2 = 8/1`