Question

The velocity of particle A is 0.1 m/sec and that of particle B is 0.05 m/sec. If the mass of particle B is 5 times that of particle A, then the ratio of de-Broglie wavelength associated with particle A and B:

• A. 2:5
• B. 3:4
• C. 6:4
• D. 5:2

Answer 1

Answer: (D)

5:2

Answer 2

The de-Broglie wavelength ($\lambda$) associated with a particle is given by the formula:

$$\lambda = \frac{h}{mv}$$

where $h$ is Planck's constant, $m$ is the mass of the particle, and $v$ is its velocity.

Given:

Velocity of particle A, $v_A = 0.1 \, \text{m/sec}$ Velocity of particle B, $v_B = 0.05 \, \text{m/sec}$

Let the mass of particle A be $m_A$. The mass of particle B is 5 times that of particle A, so $m_B = 5m_A$.

Now, we can write the de-Broglie wavelengths for particle A and particle B:

For particle A:

$$\lambda_A = \frac{h}{m_A v_A}$$

For particle B:

$$\lambda_B = \frac{h}{m_B v_B}$$

To find the ratio of de-Broglie wavelength associated with particle A and B, we divide $\lambda_A$ by $\lambda_B$:

$$\frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{m_A v_A}}{\frac{h}{m_B v_B}}$$ $$\frac{\lambda_A}{\lambda_B} = \frac{h}{m_A v_A} \times \frac{m_B v_B}{h}$$

The Planck's constant ($h$) cancels out:

$$\frac{\lambda_A}{\lambda_B} = \frac{m_B v_B}{m_A v_A}$$

Now, substitute the given values and the relationship between masses:

$$\frac{\lambda_A}{\lambda_B} = \frac{(5m_A) \times (0.05)}{m_A \times (0.1)}$$

The mass $m_A$ cancels out:

$$\frac{\lambda_A}{\lambda_B} = \frac{5 \times 0.05}{0.1}$$ $$\frac{\lambda_A}{\lambda_B} = \frac{0.25}{0.1}$$

To simplify, multiply the numerator and denominator by 100:

$$\frac{\lambda_A}{\lambda_B} = \frac{25}{10}$$

Divide both by 5:

$$\frac{\lambda_A}{\lambda_B} = \frac{5}{2}$$

Thus, the ratio of de-Broglie wavelength associated with particle A and B is 5:2.