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

Answer: (D)

$4: 1$

Answer 2

The correct option is (D) : 4:1
The kinetic energy gained by a charged particle accelerated by a potential V is qV
KE=qV
⇒2mp2​=qV⇒p=$\sqrt{2mqV}$​
p=λh​, 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