# 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 ̇.