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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.
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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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