Question

A particle of mass \(1\times10^{-30}\ \text{kg}\) and electric charge \(1.6\times10^{-19}\ \text{C}\) has de-Broglie wavelength \(660\ \text{nm}\). Then kinetic energy of this particle is
\[ (\text{Planck's constant, } h=6.6\times10^{-34}\ \text{J s}) \]

• A. \(4.2\times10^{-6}\ \text{eV}\)
• B. \(2.5\times10^{-6}\ \text{eV}\)
• C. \(1.3\times10^{-6}\ \text{eV}\)
• D. \(3.1\times10^{-6}\ \text{eV}\)

Hint:

For de-Broglie wavelength problems, \[ \lambda=\frac{h}{p} \] and \[ K=\frac{p^2}{2m} =\frac{h^2}{2m\lambda^2}. \] This direct relation is very useful in objective questions.

Answer 1

The Correct Option is D

Step 1: Find momentum from de Broglie wavelength.
\( p = \frac{h}{\lambda} = \frac{6.6\times10^{-34}}{660\times10^{-9}} = 10^{-27}\text{ kg m/s} \).

Step 2: Kinetic energy.
\( KE = \frac{p^2}{2m} = \frac{(10^{-27})^2}{2\times10^{-30}} = \frac{10^{-54}}{2\times10^{-30}} = 5\times10^{-25}\text{ J} \).
In eV: \( \frac{5\times10^{-25}}{1.6\times10^{-19}} \approx 3.1\times10^{-6}\text{ eV} \)

\[ \boxed{KE \approx 3.1\times10^{-6}\text{ eV}} \]