Question

If the momentum of an α-particle is half that of a proton, then the ratio between the wavelengths of their de-Broglie waves is

• A. 1:2
• B. 4:1
• C. 1:4
• D. 2:1
• E. 1:1

Answer 1

The Correct Option is D

The de-Broglie wavelength (\( \lambda \)) of a particle is given by the equation:

\[ \lambda = \frac{h}{p} \]

Where: - \( h \) is Planck's constant, - \( p \) is the momentum of the particle.

We are given that the momentum of the α-particle is half that of the proton. Let's denote the momentum of the proton as \( p_{\text{proton}} \) and the momentum of the α-particle as \( p_{\alpha} = \frac{1}{2} p_{\text{proton}} \).

The de-Broglie wavelength for the proton is:

\[ \lambda_{\text{proton}} = \frac{h}{p_{\text{proton}}} \]

The de-Broglie wavelength for the α-particle is:

\[ \lambda_{\alpha} = \frac{h}{p_{\alpha}} = \frac{h}{\frac{1}{2} p_{\text{proton}}} = \frac{2h}{p_{\text{proton}}} \]

Thus, the ratio of the wavelengths of the de-Broglie waves of the α-particle and the proton is:

\[ \frac{\lambda_{\alpha}}{\lambda_{\text{proton}}} = \frac{\frac{2h}{p_{\text{proton}}}}{\frac{h}{p_{\text{proton}}}} = 2 \]

Correct Answer:

Correct Answer: (D) 2:1

Answer 2

Given: 
Momentum of α-particle = \( \frac{1}{2} \) × momentum of proton

Step 1: Use de-Broglie wavelength formula: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck’s constant and \( p \) is momentum.

Let \( p_p \) be momentum of proton and \( p_\alpha = \frac{1}{2}p_p \) Then the wavelengths are: \[ \lambda_p = \frac{h}{p_p}, \quad \lambda_\alpha = \frac{h}{p_\alpha} = \frac{h}{\frac{1}{2}p_p} = \frac{2h}{p_p} \]

Step 2: Take the ratio: \[ \frac{\lambda_\alpha}{\lambda_p} = \frac{2h/p_p}{h/p_p} = 2 \Rightarrow \lambda_\alpha : \lambda_p = 2 : 1 \]

Final Answer: \( \boxed{2:1} \)