Question

De-Broglie wavelength of $\alpha$ particle is the same as that of a proton moving with (1/10) of velocity as the speed of light. The ratio of K.E. of proton to that of $\alpha$ particle is:

• A. 2:1
• B. 1:4
• C. 4:3
• D. 4:1

Hint:

Key concepts:
1. De-Broglie wavelength \( \lambda = h/p \) relates wavelength to momentum
2. For same wavelength, heavier particles move slower
3. Kinetic energy ratio depends on both mass and velocity squared
4. Always check if relativistic corrections are needed (not required here since \( v \ll c \))

Answer 1

The Correct Option is D

Given: - De-Broglie wavelength is same for both particles: \( \lambda_p = \lambda_\alpha \)
- Proton velocity \( v_p = \frac{c}{10} \)
- Mass of proton \( m_p \)
- Mass of $\alpha$-particle \( m_\alpha = 4m_p \) (since $\alpha$-particle is helium nucleus)
- Charge of $\alpha$-particle \( q_\alpha = 2e \)
Step 1: De-Broglie wavelength relation \[ \lambda = \frac{h}{p} = \frac{h}{mv} \] Since \( \lambda_p = \lambda_\alpha \): \[ \frac{h}{m_p v_p} = \frac{h}{m_\alpha v_\alpha} \] \[ m_p v_p = m_\alpha v_\alpha \] \[ v_\alpha = \frac{m_p v_p}{m_\alpha} = \frac{m_p (c/10)}{4m_p} = \frac{c}{40} \]
Step 2: Kinetic energy expressions
For non-relativistic speeds (since \( v \ll c \)): \[ KE = \frac{1}{2}mv^2 \] For proton: \[ KE_p = \frac{1}{2}m_p v_p^2 = \frac{1}{2}m_p \left(\frac{c}{10}\right)^2 = \frac{m_p c^2}{200} \] For $\alpha$-particle: \[ KE_\alpha = \frac{1}{2}m_\alpha v_\alpha^2 = \frac{1}{2}(4m_p)\left(\frac{c}{40}\right)^2 = \frac{4m_p c^2}{3200} = \frac{m_p c^2}{800} \]
Step 3: Calculate ratio
\[ \frac{KE_p}{KE_\alpha} = \frac{\frac{m_p c^2}{200}}{\frac{m_p c^2}{800}} = \frac{800}{200} = 4 \] Thus, the ratio is \( 4:1 \).