Question

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

• A. 0.25 Å
• B. 0.5 Å
• C. 1.5 Å
• D. 2 Å

Answer 1

Answer: (B)

0.5 Å

Answer 2

By using $\lambda \propto \frac{1}{\sqrt{V}}$ ⇒ $\frac{\lambda_{1}}{\lambda_{2}} = \sqrt{\frac{V_{2}}{V_{1}}}$

⇒ $\frac{10^{- 10}}{\lambda_{2}} = \sqrt{\frac{600}{150}} = 2$ ⇒ λ₂ = 0.5 Å.