Question

The de Broglie wavelength for an electron accelerated through the potential difference of \(V_1\) volt is \( \lambda_1 \). When the potential difference is changed to \(V_2\) volt, the associated de Broglie wavelength is increased by \(50%\). If \( \left(\frac{V_1}{V_2}\right)=\frac{9}{\alpha} \), then the value of \( \alpha \) is _____.

Answer 1

Correct Answer: 4

Concept: For an electron accelerated through potential \(V\): \[ \lambda=\frac{h}{\sqrt{2meV}} \] Thus \[ \lambda \propto \frac{1}{\sqrt{V}} \] Step 1: {Use wavelength relation.} Given wavelength increases by \(50%\): \[ \lambda_2=1.5\lambda_1 \] \[ \frac{\lambda_2}{\lambda_1}=1.5 \] \[ \sqrt{\frac{V_1}{V_2}}=1.5 \] Step 2: {Square both sides.} \[ \frac{V_1}{V_2}=(1.5)^2 \] \[ =\frac94 \] Step 3: {Compare with given relation.} \[ \frac{V_1}{V_2}=\frac{9}{\alpha} \] Thus \[ \frac{9}{\alpha}=\frac94 \] \[ \alpha=4 \]