Question

The ratio of wavelengths of proton and deuteron accelerated by potential $V_p$ and $V_d$ is $1: \sqrt{2}$ Then, the ratio of $V_p$ to $V_d$ will be

• A. $1: 1$
• B. $\sqrt{2}: 1$
• C. $2: 1$
• D. $4: 1$

Answer 1

The Correct Option is D

The correct option is (D) : 4:1 
The kinetic energy gained by a charged particle accelerated by a potential V is qV 
K_E=qV 
⇒2mp^2​=qV⇒p=\(\sqrt{2mqV}\)​ 
p=λ_h​, thus λ=\(\frac{h}{\sqrt{2mqV}}\)
now \(\frac{\lambda_p}{\lambda_d}\)=\(\sqrt{\frac{m_dV_d}{M_pV_p}}\)
\(\Rightarrow\)\(\frac{1}{\sqrt2}\)=\(\sqrt{\frac{2V_d}{V_p}}\)\(\Rightarrow\)\(\frac{V_p}{V_d}\)=4