Question

The angular momentum of electron in H atom in \(n_x\) state is \(1.051\times10^{-34}\,\mathrm{J\,s}\). The de Broglie wavelength of electron in this \(n_x\) state is \[ (h=6.6\times10^{-34}\,\mathrm{J\,s};\ \pi=3.14) \]

• A. \(33.2\times10^{-2}\,\mathrm{nm}\)
• B. \(66.4\times10^{-2}\,\mathrm{nm}\)
• C. \(332.1\times10^{-2}\,\mathrm{nm}\)
• D. \(662.4\times10^{-2}\,\mathrm{nm}\)

Hint:

For Bohr's model, \[ \boxed{ L=n\frac{h}{2\pi} } \] and \[ \boxed{ 2\pi r_n=n\lambda. } \]

Answer 1

The Correct Option is A

Step 1: Find the principal quantum number. According to Bohr's quantization, \[ L=n\frac{h}{2\pi}. \] Given, \[ L=1.051\times10^{-34}\,\mathrm{J\,s}. \] Thus, \[ n = \frac{2\pi L}{h} = \frac{2(3.14)(1.051\times10^{-34})} {6.6\times10^{-34}} \approx1. \] Hence, \[ n=1. \]

Step 2: Use the de Broglie relation. For the hydrogen atom, \[ 2\pi r_n=n\lambda. \] For the first orbit, \[ r_1=0.0528\,\mathrm{nm}. \] Therefore, \[ \lambda = 2\pi r_1 = 2(3.14)(0.0528) \approx0.332\,\mathrm{nm}. \] Thus, \[ \lambda = 33.2\times10^{-2}\,\mathrm{nm}. \] Hence, \[ \boxed{33.2\times10^{-2}\,\mathrm{nm}} \] Therefore, \[ \boxed{(A)} \] is the correct answer.