A proton accelerated through a potential difference V has a de Broglie wavelength \( \lambda \). On doubling the accelerating potential, de Broglie wavelength of the proton ____________________________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → decreases
A proton of charge \( e \) accelerated through \( V \) gains kinetic energy \( eV \), so its momentum is \( p = \sqrt{2 m e V} \). The de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m e V}} \;\Rightarrow\; \lambda \propto \frac{1}{\sqrt{V}} \)
Doubling the potential:
\( \displaystyle \lambda' = \frac{\lambda}{\sqrt{2}} \approx 0.71\,\lambda \)
So the wavelength decreases.
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{m v} \qquad (11.5) \)
where m is the mass of the particle and v its speed. Equation (11.5) is known as the de Broglie relation and the wavelength λ of the matter wave is called de Broglie wavelength.”