A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV . Find the energy and wavelength of photon emitted.

Explanation

Let `n_(1)` be initial state of electron. Then `E_(1) = - (13.6)/(n_(1)^(2))`eV here `E_(1) `= - 0.85 eV, therefore - 0.85 = `- (13.6)/(n_(1)^(2)) " or " n_(1) = 4 `
(ii) Let `n_(2)` be the final excitation state of the electron. Since excitation energy is always measured with respect to the ground state, therefore
`Delta` E = 13.6 ` [ 1- (1)/(n_(2)^(2)) ] ` here `Delta`E = 10.2 eV, therefore
10.2 = 13.6 `[ 1 - (1)/(n_(2)^(2)) ] " or " n_(2) = 2 `
Thus, the electron jumps from `n_(1) `= 4 to `n_(1) = ` .
(iii) The wavelength of the photon emitted for a transition between `n_(1)` = 4 to `n_(2)` = 2, is given by
`(1)/(lambda) = R_(infty) [ (1)/(n_(2)^(2)) - (1)/(n_(1)^(2)) ] " (or)" (1)/(lambda) = 1.09 xx 10^(7) [ (1)/(2^(2)) - (1)/(4^(2)) ] `
`lambda = 4860 `Å