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. What is the initial state of hydrogen atom ?

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 n₁ be initial state of electron.

Then E₁ = -13.6/(n₁² ) eV ; Here E₁ = -0.85 ev,

Therefore , -0.85 = 13.6/(n₁² )

or, n₁ = 4.

(ii) Let n₂ 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/(n₂² )]

here , ΔE = 10.2 eV, therefore, 10.2 = 13.6 [1 – 1/(n₂² )]

or, n₂ = 2

Thus the electron jumps from n₁ = 4 to n₂ = 2.

(iii) The wavelength of the photon emitted for a transition between n₁ = 4 to n₂ = 2,is given by

1/( λ) = R [1/(n₂² )– 1/(n₁² )]

(or) 1/λ=1.09 × 10⁷ [1/2² – 1/4² ]= 4860 A ̇.