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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.
Solution
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
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