Question

The kinetic energy of electron and proton is$10^{- 32}J$. Then the relation between their de-Broglie wavelengths is

• A. $\lambda_{p} < \lambda_{e}$
• B. $\lambda_{p} > \lambda_{e}$
• C. $\lambda_{p} = \lambda_{e}$
• D. $\lambda_{p} = 2\lambda_{e}$

Answer 1

Answer: (A)

$\lambda_{p} < \lambda_{e}$

Answer 2

By using $\lambda = \frac{h}{\sqrt{2mE}}$ E = 10^–32 J = Constant for both particles. Hence $\lambda \propto \frac{1}{\sqrt{m}}$

Since $m_{p} > m_{e}$ so $\lambda_{p} < \lambda_{e}.$