Question

When an electron at rest is accelerated through an electric potential, the de Broglie wavelength associated with the electron is \(\lambda\). For the de Broglie wavelength associated with the electron to become \(\frac{2\lambda}{3}\), the percentage increase in the potential to be applied is:

• A. \(75\)
• B. \(225\)
• C. \(125\)
• D. \(150\)

Hint:

Remember the important relation \(\lambda\propto 1/\sqrt{V}\). A decrease in wavelength requires an increase in accelerating potential.

Answer 1

The Correct Option is C

Concept: For an electron accelerated through a potential difference \(V\), \[ \lambda=\frac{h}{\sqrt{2meV}} \] Hence, \[ \lambda\propto\frac{1}{\sqrt{V}} \] Therefore, \[ \lambda\sqrt{V}=\text{constant} \]

Step 1: Relate the two wavelengths. Initially, \[ \lambda_1=\lambda \] Finally, \[ \lambda_2=\frac{2\lambda}{3} \] Using \[ \lambda_1\sqrt{V_1} = \lambda_2\sqrt{V_2} \] we get \[ \lambda\sqrt{V_1} = \frac{2\lambda}{3}\sqrt{V_2} \] \[ \sqrt{V_2} = \frac{3}{2}\sqrt{V_1} \] Squaring, \[ V_2 = \frac{9}{4}V_1 \]

Step 2: Calculate percentage increase. Increase in potential: \[ \Delta V = V_2-V_1 = \left(\frac94-1\right)V_1 \] \[ = \frac54V_1 \] Percentage increase: \[ \frac{\Delta V}{V_1}\times100 = \frac54\times100 \] \[ =125\% \] \[ \boxed{125\%} \]