Question

The velocity of the electron in Bohr’s first orbit is \( x \times 10^6 \, \text{ms}^{-1} \). The deBroglie wavelength associated with it (in nm) is (Given: \( m_e = 9 \times 10^{-31} \, \text{kg}, \, h = 6.6 \times 10^{-34} \, \text{Js} \))

• A. \( \frac{x}{1.43} \)
• B. \( \frac{x}{0.73} \)
• C. 0.73
• D. \( \frac{0.073}{x} \)

Hint:

Use de Broglie’s equation \( \lambda = \frac{h}{mv} \) and convert to nanometers.
Always double-check the unit consistency—especially when switching from m to nm.

Answer 1

The Correct Option is C

Using de Broglie relation: \[ \lambda = \frac{h}{mv} \] Given: \[ m = 9 \times 10^{-31} \, \text{kg}, \quad v = x \times 10^6 \, \text{m/s}, \quad h = 6.6 \times 10^{-34} \, \text{Js} \] Substitute: \[ \lambda = \frac{6.6 \times 10^{-34}}{9 \times 10^{-31} \cdot x \times 10^6} = \frac{6.6}{9x} \times 10^{-9} \, \text{m} = \frac{0.73}{x} \, \text{nm} \] So, when \( x = 1 \), the wavelength is 0.73 nm.