Question

If the de-Broglie wavelength is $\lambda _{0}$ for protons accelerated through $100 \, V$ , then the de-Broglie wavelength for alpha particles accelerated through the same voltage will be

• A. $\lambda _{0}$
• B. $\frac{\lambda _{0}}{2}$
• C. $\frac{\lambda _{0}}{2 \sqrt{2}}$
• D. none of these

Answer 1

Answer: (C)

De-Broglie wavelength is $\lambda =\frac{h}{\sqrt{2 m K}}=\frac{h}{\sqrt{2 m q V}}$
Therefore, de-Broglie wavelength for protons.
$\left(\lambda \right)_{0}=\frac{h}{\sqrt{2 \left(m_{p}\right) e V_{0}}}$
and de-Broglie wavelength for alpha particles
$\left(\lambda \right)^{′}=\frac{h}{\sqrt{2 \left(4 m_{p}\right) \left(2 e\right) V_{0}}}$
$\frac{\lambda _{0}}{\lambda ^{′}}=\sqrt{8}=2\sqrt{2}$
$\lambda ^{′}=\frac{\lambda _{0}}{2 \sqrt{2}}$