Question

An $\alpha$ particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are $\lambda_\alpha$ and $\lambda_p$ respectively. The ratio $\lambda_p/\lambda_\alpha$ is :

• A. 2.8
• B. 8
• C. 7.8
• D. 3.8

Hint:

When particles are accelerated by the {same potential}, the wavelength ratio depends only on the square root of the product of their mass and charge: $\lambda \propto \frac{1}{\sqrt{mq}}$.

Answer 1

The Correct Option is A

To find the ratio of de Broglie wavelengths of a proton and an α particle, we will use the concept of de Broglie wavelength, which is given by:

\(\lambda = \frac{h}{p}\)

where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

For particles accelerated through a potential difference \(V\), the kinetic energy \(K.E.\) is given by:

\(K.E. = \frac{1}{2}mv^2 = qV\)

where \(q\) is the charge of the particle and \(v\) is its velocity.

The momentum \(p\) can be expressed as:

\(p = mv = \sqrt{2mqV}\)

Thus, the de Broglie wavelength \(\lambda\) is:

\(\lambda = \frac{h}{\sqrt{2mqV}}\)

For an α particle (which is a helium nucleus), it has a charge \(q_\alpha = 2e\) and a mass of approximately \(4m_p\), where \(m_p\) is the mass of a proton.

The de Broglie wavelength for the α particle is: 

\(\lambda_\alpha = \frac{h}{\sqrt{2 \times 4m_p \times 2e \times V}}\) = \frac{h}{\sqrt{16m_peV}}

For a proton, the charge \(q_p = e\) and mass \(m = m_p\). Its de Broglie wavelength is:

\(\lambda_p = \frac{h}{\sqrt{2m_peV}}\)

The ratio of the wavelengths is:

\(\frac{\lambda_p}{\lambda_\alpha} = \frac{\frac{h}{\sqrt{2m_peV}}}{\frac{h}{\sqrt{16m_peV}}} = \sqrt{\frac{16m_peV}{2m_peV}} = \sqrt{8} = 2.8\)

Therefore, the correct answer is 2.8.