A proton is accelerated by a potential V and the de-Broglie wavelength associated with it is \( \lambda \). By how much potential a deuteron 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) → \( V/2 \)
A charged particle accelerated through a potential \( V \) gains kinetic energy \( qV \), so its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}} \)
For the same \( \lambda \) the product \( mqV \) must be the same. A deuteron has the same charge as a proton (\( q_d = q_p = e \)) and about twice the mass (\( m_d = 2m_p \)).
\( \displaystyle m_p\,V = m_d\,V_d \;\Rightarrow\; V_d = \frac{m_p}{m_d}\,V = \frac{V}{2} \)
The deuteron needs only half the potential, because its greater mass gives it a larger momentum at the same energy.