Question

The de-Broglie wavelength of a particle accelerated with $150$ volt potential is $10^{-10}\, m$. If it is accelerated by $600$ volts p.d., its wavelength will be

• A. $0.25\, \mathring{A}$
• B. $0.5\, \mathring{A}$
• C. $1.5\, \mathring{A}$
• D. $2\,\mathring{A}$

Answer 1

Answer: (B)

By using $\lambda \propto \frac{1}{\sqrt{V}}$
$\Rightarrow \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{V_{2}}{V_{1}}}$
$\Rightarrow \frac{10^{-10}}{\lambda_{2}}=\sqrt{\frac{600}{150}}=2$
$\Rightarrow \lambda_{2}=0.5\, \mathring{A}$