Question

The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

• A. \(0.529\ \mathring{\mathrm{A}}\)
• B. \(2\pi \times 0.529\ \mathring{\mathrm{A}}\)
• C. \(\frac{0.529}{2\pi}\ \mathring{\mathrm{A}}\)
• D. \(4 \times 0.529\ \mathring{\mathrm{A}}\)

Hint:

Bohr's quantisation \(mvr = nh/2\pi\) with \(\lambda = h/mv\) gives \(2\pi r = n\lambda\); for n = 1, \(r = 0.529\ \mathring{\mathrm{A}}\).

Answer 1

The Correct Option is B

Bohr's condition: \(mvr = \dfrac{nh}{2\pi}\).
de Broglie: \(\lambda = \dfrac{h}{mv}\), so \(mv = \dfrac{h}{\lambda}\).
Substituting: \(\dfrac{h}{\lambda}r = \dfrac{nh}{2\pi}\), hence \(\lambda = \dfrac{2\pi r}{n}\).
For the first orbit n = 1 and \(r = 0.529\ \mathring{\mathrm{A}}\): \(\lambda = 2\pi \times 0.529\ \mathring{\mathrm{A}} \approx 3.32\ \mathring{\mathrm{A}}\).
Answer: (2) \(2\pi \times 0.529\ \mathring{\mathrm{A}}\)