Question

Two particles A and B are in motion and the wavelength of A is 5 × 10^-8 m. The wavelength of B, so that its momentum is double that of A, will be:

• (A) 3 × 10^-4 m
• (B) 2.5 × 10^-8 m
• (C) 2.5 × 10^-6 m
• (D) 1.6 × 10^-8 m

Answer 1

The Correct Option is B

The question involves calculating the wavelength of particle B, given that particle A has a certain wavelength and the momentum of B is double that of A. This can be solved using the de Broglie wavelength formula, which relates a particle's wavelength to its momentum.

The formula for the de Broglie wavelength \lambda is given by:

\lambda = \frac{h}{p}

where:

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

Given that the wavelength of particle A, \lambda_A, is 5 \times 10^{-8} m, we use the formula:

\lambda_A = \frac{h}{p_A}

From this, we rearrange to find the momentum of particle A:

p_A = \frac{h}{\lambda_A}

Now, we are told that the momentum of particle B, p_B, is double that of particle A:

p_B = 2p_A = 2 \cdot \frac{h}{\lambda_A}

We need to find the wavelength of particle B, \lambda_B:

\lambda_B = \frac{h}{p_B} = \frac{h}{2p_A}

Substituting for p_A:

\lambda_B = \frac{h}{2 \cdot \frac{h}{\lambda_A}} = \frac{\lambda_A}{2}

Now plug in the value of \lambda_A = 5 \times 10^{-8} m:

\lambda_B = \frac{5 \times 10^{-8}}{2} = 2.5 \times 10^{-8} m

Thus, the wavelength of particle B is 2.5 \times 10^{-8} m, making the correct answer:

(B) 2.5 × 10^-8 m