Question

An electron accelerated by a potential difference '$V$' has de-Broglie wavelength '$\lambda$'. If the electron is accelerated by a potential difference '$9 \text{ V}$', its de-Broglie wavelength will be}

• A. $\frac{\lambda}{4.5}$
• B. $\frac{\lambda}{3}$
• C. $\frac{\lambda}{2}$
• D. $\lambda$

Hint:

Potential increases by 9 times $\rightarrow$ Wavelength decreases by $\sqrt{9} = 3$ times.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The de-Broglie wavelength of a charged particle relates to its momentum.
When accelerated through a potential difference, electrical potential energy converts to kinetic energy, determining the momentum.
Step 2: Key Formula or Approach:
Kinetic energy $K = eV$.
Momentum $p = \sqrt{2mK} = \sqrt{2meV}$.
de-Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.
Thus, $\lambda \propto \frac{1}{\sqrt{V}}$.
Step 3: Detailed Explanation:
Let the initial wavelength be $\lambda$ for potential $V$: \[ \lambda = \frac{h}{\sqrt{2meV}} \] Let the new wavelength be $\lambda'$ for potential $V' = 9V$.
Using the proportionality: \[ \frac{\lambda'}{\lambda} = \sqrt{\frac{V}{V'}} \] Substitute $V' = 9V$: \[ \frac{\lambda'}{\lambda} = \sqrt{\frac{V}{9V}} \] \[ \frac{\lambda'}{\lambda} = \sqrt{\frac{1}{9}} \] \[ \frac{\lambda'}{\lambda} = \frac{1}{3} \] Rearrange to find the new wavelength: \[ \lambda' = \frac{\lambda}{3} \] Step 4: Final Answer:
The new wavelength is $\frac{\lambda}{3}$.