Question

An electron is accelerated from rest by 480 V, its wavelength is $\lambda$. If accelerated by 120 V, the wavelength is:

• A. $5\lambda$
• B. $4\lambda$
• C. $2\lambda$
• D. $3\lambda$
• E. $6\lambda$

Hint:

If voltage decreases by a factor of 4, the wavelength increases by a factor of $\sqrt{4} = 2$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem deals with the de Broglie wavelength of an electron that has been accelerated through a potential difference. We need to find the relationship between the accelerating potential and the resulting de Broglie wavelength.
Step 2: Key Formula or Approach:
1. Kinetic Energy: An electron (charge e) accelerated from rest through a potential difference V gains a kinetic energy (K) of \( K = eV \). 2. de Broglie Wavelength: The wavelength (\( \lambda \)) of a particle is related to its momentum (p) by \( \lambda = h/p \), where h is Planck's constant. 3. Relating KE and Momentum: Kinetic energy is related to momentum by \( K = p^2 / (2m) \), so \( p = \sqrt{2mK} \). 4. Combining the formulas: Substitute (1) and (3) into (2) to get the de Broglie wavelength in terms of the accelerating potential V: \[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2meV}} \] From this final formula, we can see the relationship between wavelength and potential: \[ \lambda \propto \frac{1}{\sqrt{V}} \] Step 3: Detailed Explanation:
We have two scenarios. Let \( \lambda_1 \) be the wavelength for potential \( V_1 \) and \( \lambda_2 \) be the wavelength for potential \( V_2 \). From the proportionality \( \lambda \propto 1/\sqrt{V} \), we can write the ratio: \[ \frac{\lambda_2}{\lambda_1} = \frac{1/\sqrt{V_2}}{1/\sqrt{V_1}} = \sqrt{\frac{V_1}{V_2}} \] We are given: - \( V_1 = 480 \text{ V} \), \( \lambda_1 = \lambda \) - \( V_2 = 120 \text{ V} \), \( \lambda_2 = ? \) Substitute the values into the ratio: \[ \frac{\lambda_2}{\lambda} = \sqrt{\frac{480}{120}} \] \[ \frac{\lambda_2}{\lambda} = \sqrt{4} = 2 \] Solve for \( \lambda_2 \): \[ \lambda_2 = 2\lambda \] Step 4: Final Answer:
The new wavelength is \( 2\lambda \).