The electron, in a hydrogen atom, initially in a state of quantum number n1

makes a transition to a state whose excitation energy, with respect to the

ground state, is 10.2 eV. If the wavelength, associated with the photon emitted

in this transition, is 487.5 mm, find the

(i) energy in ev, and (ii) value of the quantum number, n1 of the electron in its

initial state. 

for a  hydrogen atom, the energy in terms of principal quantum number 'n' is given as
E_n = (-13.6/n²) eV

therefore

for n = 1
E₁ = -13.6 / 1² =  -13.6 eV 

and n = 2

E₂ = -13.6 / 2² =  -3.4 eV

Now,
as per given for state n=2 has the excitation energy of the atom is 10.2 eV. So, the electron is making a transition from n = n₁ to n=2 where (n₁>2).
Now, we have 

E_n1 – E₂ = hc/λ

here

 hc/λ = [(6.63 x 10^-34 x 3 x 10⁸ ) / (487.5 x 10^-3)] x (1/ 1.6 x 10^-19) eV

or hc/λ = 2.55 eV
now as

E_n1= E2 + hc/λ =  (-3.4 + 2.55) eV
so,

E_n1 ≃ - 0.85 eV
But we also have E_n1 =  (-13.6/n₁²) eV
or

n₁ = (-13.6 /E_n1 )^1/2 = (-13.6 / -0.85)^1/2

thus, we have

n₁ = 4