A proton is accelerated by a potential difference of \( V \) Volt. Its de Broglie wavelength is \( \lambda \). If an alpha particle is accelerated by a potential difference of \( 2V \) Volt, then de Broglie wavelength of the alpha particle will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{\lambda}{4} \)
A charge \( q \) accelerated from rest through a potential difference \( V \) gains kinetic energy \( qV \), so its momentum is \( \sqrt{2mqV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{\sqrt{2mqV}} \)
An alpha particle has four times the mass and twice the charge of a proton: \( m_{\alpha} = 4m \), \( q_{\alpha} = 2e \), and here \( V_{\alpha} = 2V \).
\( \displaystyle \frac{\lambda_{\alpha}}{\lambda} = \sqrt{\frac{m\,e\,V}{(4m)(2e)(2V)}} = \sqrt{\frac{1}{16}} = \frac{1}{4} \)
\( \displaystyle \lambda_{\alpha} = \frac{\lambda}{4} \)
Note: NTA picked this question from an NCERT Exemplar problem and changed the figures. Do not ignore the Exemplar while practising numericals.
NCERT Exemplar:
Dual Nature of Radiation and Matter – …
