Question

Calculate the de-Broglie wavelength of the electron orbitting in the n = 2 state of hydrogen atom.

Answer 1

As we know,

`mvr_2=(nh)/(2pi)`

The de-Broglie wavelength is given as,

`lambda=h/(mv)=h/(nh)xx(2pir_2)=pir_2`

As r₂=0.529×n^₂×10⁻¹⁰ m

Therefore, λ=π×0.529×10⁻¹⁰×n²=3.14×0.529×10⁻¹⁰×2²≃6.64×10⁻¹⁰ m