Question

If the kinetic energy (K.E.) of an electron is $2.5\times {10}^{–24}\text{J}$, calculate its de-Broglie wavelength.

• A. 400 nm
• B. 380 nm
• C. 311 nm
• D. 298 nm

Answer 1

The correct option is C 311 nm

Kinetic energy (K.E.) $=\frac{1}{2}m{\text{v}}^2$
$\Rightarrow \text{v}=\sqrt{\frac{2K.E.}{m}}$

Since, mass of an electron is $9.1\times {10}^{–31}\text{kg}$
Substituting the values, we get
$\Rightarrow \text{v}=\sqrt{\frac{2\times 2.5\times {10}^{–24}\text{J}}{9.1\times {10}^{–31}\text{kg}}}=2.34\times {10}^3{\text{m s}}^{-1}$

$\text{Wavelength}\left(\lambda \right)=\frac{h}{m\text{v}}$
$=\frac{6.626\times {10}^{-34}\text{J s}}{(9.1\times {10}^{–31}\text{kg})(2.34\times {10}^3{\text{m s}}^{-1})}=311.1\times {10}^{–9}\text{m}$

On solving, $\lambda =311.1\text{nm}$