Question

The ratio of wavelengths of proton and deuteron accelerated by potential $V_p$ and $V_d$ is $1: \sqrt{2}$ Then, the ratio of $V_p$ to $V_d$ will be

• A. $1: 1$
• B. $\sqrt{2}: 1$
• C. $2: 1$
• D. $4: 1$

Answer 1

The Correct Option is D

To solve the problem, we need to understand the relationship between the wavelength and the accelerating potential for both a proton and a deuteron. The de Broglie wavelength (\lambda) for a particle is given by:

\lambda = \frac{h}{\sqrt{2mE_k}}

where:

• h is the Planck's constant.
• m is the mass of the particle.
• E_k is the kinetic energy of the particle, which can be related to the potential V applied as E_k = qV, where q is the charge of the particle.

Both proton and deuteron have the same charge (q = e), but different masses. The mass of a proton (m_p) and the mass of a deuteron (m_d = 2m_p) are related by the factor of 2.

The ratio of wavelengths for proton and deuteron is given as:

\frac{\lambda_p}{\lambda_d} = 1:\sqrt{2}

Substituting the expressions for the wavelengths, we have:

\frac{\frac{h}{\sqrt{2m_p e V_p}}}{\frac{h}{\sqrt{2m_d e V_d}}} = \frac{\sqrt{2m_d e V_d}}{\sqrt{2m_p e V_p}} = 1:\sqrt{2}

Simplifying, we get:

\frac{\sqrt{m_d V_d}}{\sqrt{m_p V_p}} = 1:\sqrt{2}

Squaring both sides gives:

\frac{m_d V_d}{m_p V_p} = \frac{1}{2}

Substitute m_d = 2m_p:

\frac{2m_p V_d}{m_p V_p} = \frac{1}{2}

After simplifying:

\frac{V_d}{V_p} = \frac{1}{4}

This implies:

\frac{V_p}{V_d} = 4:1

Thus, the ratio V_p to V_d is 4:1, making the correct option:

Option: 4:1