Question

An electron and a proton has same de Broglie wavelength. If K_e and K_p are their respective kinetic energies, then

• A. k_p > k_e
• B. k_p < k_e
• C. k_p = k_e
• D. None of these

Answer 1

The Correct Option is B

To solve this problem, we need to compare the kinetic energies of an electron and a proton when they have the same de Broglie wavelength.

The de Broglie wavelength \(\lambda\) for a particle is given by the formula: 

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

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

Since the electron and the proton have the same de Broglie wavelength, their momenta are equal:

\(\frac{h}{p_e} = \frac{h}{p_p}\).

The momentum \(p\) of a particle is also related to its mass \(m\) and velocity \(v\) by:

\(p = mv\)

Therefore, if the momenta are equal, we have:

\(m_e v_e = m_p v_p\) (1)

The kinetic energy \(K\) of a particle is given by:

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

For the electron, the kinetic energy \(K_e\) is:

\(K_e = \frac{1}{2} m_e v_e^2\) (2)

For the proton, the kinetic energy \(K_p\) is:

\(K_p = \frac{1}{2} m_p v_p^2\) (3)

From equation (1) \((m_e v_e = m_p v_p)\), expressing the velocity in terms of momentum:

\(v_e = \frac{p_e}{m_e}\) and \(v_p = \frac{p_p}{m_p}\).

Substituting in the kinetic energy formulae:

\(\Rightarrow K_e = \frac{1}{2} m_e \left(\frac{p_e}{m_e}\right)^2 = \frac{p_e^2}{2 m_e}\)

\(\Rightarrow K_p = \frac{1}{2} m_p \left(\frac{p_p}{m_p}\right)^2 = \frac{p_p^2}{2 m_p}\)

Since \(p_e = p_p\), comparing kinetic energies:

\(\frac{p^2}{2 m_e}\) vs \(\frac{p^2}{2 m_p}\).

As the mass of a proton \((m_p)\) is significantly greater than that of an electron \((m_e)\)\), it follows that:

\(\frac{1}{2 m_p} < \frac{1}{2 m_e}\)

Hence, \(K_p < K_e\).

Conclusion: Therefore, the correct answer is \(K_p < K_e\).