A proton is accelerated by a potential V and the de-Broglie wavelength associated with it is 2. By how much potential a deutron should be accelerated to acquire the same de-Broglie wavelength as that of a proton?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → V/2
de-Broglie wavelength: $\lambda = \frac{h}{p}$
For a particle accelerated by a potential $V$, kinetic energy: $K = qV = \frac{1}{2} m v^2$
Momentum: $p = \sqrt{2 m q V}$
de-Broglie wavelength: $\lambda = \frac{h}{\sqrt{2 m q V}}$
Given: proton with mass $m_p$, charge $e$, accelerated by $V_p = V$, wavelength $\lambda = 2$
For deuteron: mass $m_d = 2 m_p$, charge $e$, want same wavelength $\lambda = 2$
Using: $\lambda \propto \frac{1}{\sqrt{m V}} \Rightarrow \frac{\lambda_d}{\lambda_p} = \sqrt{\frac{m_p V_p}{m_d V_d}}$
$1 = \sqrt{\frac{m_p V}{2 m_p V_d}} = \sqrt{\frac{V}{2 V_d}} \Rightarrow 1 = \sqrt{\frac{V}{2 V_d}}$
Squaring both sides: $1 = \frac{V}{2 V_d} \Rightarrow V_d = \frac{V}{2}$
Answer: Deuteron should be accelerated by $V/2$ to have the same de-Broglie wavelength