Question

A proton and a $Li^{3+}$ nucleus are accelerated by the same potential. If $\lambda_{Li}$ and $\lambda_p$ denote the de Broglie wavelengths of $Li^{3+}$ and proton respectively, then the value of $\lambda_{Li} / \lambda_p$ is $x \times 10^{-1}$. The value of $x$ is __________. (Rounded off to the nearest integer) [Mass of $Li^{3+}$ = 8.3 mass of proton]

Hint:

When particles are accelerated through the same potential, their wavelength is inversely proportional to the square root of their mass-charge product.

Answer 1

Correct Answer: 2

Step 1: de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}}$.
Step 2: Since $V$ is the same, $\lambda \propto \frac{1}{\sqrt{mq}}$.
Step 3: Ratio $\frac{\lambda_{Li}}{\lambda_p} = \sqrt{\frac{m_p q_p}{m_{Li} q_{Li}}}$.
Step 4: Given $m_{Li} = 8.3 m_p$ and $q_{Li} = 3e$ (for $Li^{3+}$), $q_p = e$.
Step 5: $\frac{\lambda_{Li}}{\lambda_p} = \sqrt{\frac{1 \times 1}{8.3 \times 3}} = \sqrt{\frac{1}{24.9}} \approx \sqrt{0.04016} \approx 0.2004$.
Step 6: $0.2004 = 2.004 \times 10^{-1} \approx 2 \times 10^{-1}$. Thus, $x = 2$.