Question

In hydrogen atom, the de Broglie wavelength of an electron in the first Bohr's orbit is ____________.

[Given that Bohr radius, a₀ = 52.9 pm]

Options

• 105.8 π pm

• 26.45 π pm

• 211.6 π pm

• 52.9 π pm

Answer 1

In hydrogen atom, the de Broglie wavelength of an electron in the first Bohr's orbit is 105.8 π pm.

Explanation:

r = a₀n²

∴ r = 52.9 × (1)² pm

Now, nλ = 2πr

λ = `(2π"r")/"n"`

= `(2π xx 52.9 xx 1^2)/1`

= 105.8 π pm