Q: 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.

(i) What is the initial state of hydrogen atom?

(ii) What is the final state of hydrogen atom?

(iii) What is the wavelength of the photon emitted ?

Sol: (i) 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 , ΔE = 13.6 [1- 1/(n_{2}^{2} )]

here , ΔE = 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_{2} = 2.

(iii) The wavelength of the photon emitted for a transition between n_{1} = 4 to n_{2} = 2,is given by

1/( λ) = R [1/(n_{2}^{2} )– 1/(n_{1}^{2} )]

(or) 1/λ=1.09 × 10^{7} [1/2^{2} – 1/4^{2} ]= 4860 A ̇.