Question

Given the ground state energy E₀ = - 13.6 eV and Bohr radius a₀ = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.

Answer 1

Ground state energy is E₀ = –13.6 eV

Energy in the first excited state = E₁ = `(-13.6eV)/(2)^2=-3.4eV`

We know that, the de Broglie wavelength λ  is given as

λ = `h/p`

But , p = `sqrt(2mE_1)`

`:.lambda = h/sqrt(2mE_1)`

`=(6.63xx10^(-34))/sqrt(2xx(9.1xx10^(-31))xx(3.4xx1.6xx10^(-19)))`

`=(6.63xx10^(-34))/(9.95xx10^(-25)`

λ = 6.6 x 10^-10 m