Question

The radius of the 4^th Bohr orbit is 0.864 nm. The de Broglie wavelength of the electron in that orbit is –

• A. 13.297 Å
• B. 1.3565 nm
• C. 1.3291 pm
• D. 0.1329 nm

Answer 1

Answer: (B)

1.3565 nm

Answer 2

2p r_n = nl

Ž l = $\frac{2\pi r_{n}}{n} = \frac{2 \times 3.14 \times 0.864}{4}$ nm

= 1.3565 nm