Question

The wavelength associated with the electron moving in the first orbit of hydrogen atom with velocity $ 2.2 \times 10^6\, \text{m/s} $ is (in nm):
(Given: $ m_e = 9.0 \times 10^{-31}\, \text{kg},\ h = 6.6 \times 10^{-34}\, \text{Js} $)

• A. 0.66 nm
• B. 0.33 nm
• C. 0.22 nm
• D. 0.44 nm

Hint:

Use \( \lambda = \frac{h}{mv} \), and always convert final answer to the required unit (e.g., nm).

Answer 1

The Correct Option is B

We use the de Broglie wavelength formula: \[ \lambda = \frac{h}{mv} \] Substitute the given values: \[ \lambda = \frac{6.6 \times 10^{-34}}{9.0 \times 10^{-31} \cdot 2.2 \times 10^6} = \frac{6.6 \times 10^{-34}}{1.98 \times 10^{-24}} \approx 3.33 \times 10^{-10}\, \text{m} \] Convert to nanometers: \[ \lambda = 0.333\, \text{nm} \approx 0.33\, \text{nm} \]