Question

If the accelerating potential of a moving charged particle of \( \lambda \) de Broglie wavelength is increased by four times, then the de Broglie wavelength will be:

• A. \( \lambda/4 \)
• B. \( \lambda/2 \)
• C. \( 2\lambda \)
• D. \( 4\lambda \)

Hint:

Use \( \lambda = h/\sqrt{2mqV} \); the wavelength varies inversely with the square root of the accelerating potential.

Answer 1

The Correct Option is B

Step 1: The work done by the accelerating field equals the kinetic energy: \(qV = \tfrac{1}{2}mv^2\), so the speed is \(v = \sqrt{2qV/m}\).

Step 2: Momentum is \(p = mv = \sqrt{2mqV}\), and wavelength is \(\lambda = h/p\). Since \(h\), \(m\) and \(q\) are fixed, only \(V\) changes, giving \(\lambda \propto V^{-1/2}\).

Step 3: Write the ratio directly. Quadrupling \(V\) multiplies \(\sqrt{V}\) by \(\sqrt{4}=2\), and because \(\lambda\) sits in the denominator, \(\lambda\) is divided by 2.

Step 4: Hence the wavelength becomes half of the original value.

\[\boxed{\lambda' = \frac{\lambda}{2}}\]