Question

If the kinetic energies of an electron, an alpha particle, and a proton having the same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2, \varepsilon_3$ respectively, then

• A. ε1>ε3>ε2
• B. ε1 = ε2 = ε3
• C. ε1<ε3<ε2
• D. ε1>ε2>ε3

Answer 1

The Correct Option is A

Given:
The electron, alpha particle, and proton have the same de-Broglie wavelength. Let their kinetic energies be \( \varepsilon_1, \varepsilon_2, \varepsilon_3 \) respectively.

Step 1: Relate kinetic energy to de-Broglie wavelength
The de-Broglie wavelength \( \lambda \) is given by: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant, and \( p \) is the momentum of the particle. The momentum of a particle is related to its kinetic energy \( K \) by: \[ K = \frac{p^2}{2m} \] Thus, the momentum is: \[ p = \sqrt{2mK} \] Substituting this into the de-Broglie wavelength formula: \[ \lambda = \frac{h}{\sqrt{2mK}} \] Rearranging to solve for kinetic energy \( K \): \[ K = \frac{h^2}{2m\lambda^2} \] 

Step 2: Compare the kinetic energies
Since the de-Broglie wavelength is the same for all three particles, we see that the kinetic energy is inversely proportional to the mass of the particle. Therefore, the particle with the smallest mass will have the highest kinetic energy. - Mass of the electron \( m_e \) - Mass of the proton \( m_p \) - Mass of the alpha particle \( m_{\alpha} \) The masses are ordered as: \[ m_e < m_p < m_{\alpha} \] Thus, the kinetic energies will follow the order: \[ \varepsilon_1 > \varepsilon_3 > \varepsilon_2 \] where \( \varepsilon_1 \) is the kinetic energy of the electron, \( \varepsilon_3 \) is the kinetic energy of the proton, and \( \varepsilon_2 \) is the kinetic energy of the alpha particle. 

Answer: \( \varepsilon_1 > \varepsilon_3 > \varepsilon_2 \)