Answer :
The initial and final states if an electron in hydrogen emits a photon with a wavelength of 656 nm is n = 3 to n = 2.
given that :
wavelength of the photon emitted λ = 656 nm = 656 × 10⁻⁹ m
the energy of the photon is given as :
E = hc / λ
= (6.626 × 10⁻³⁴ × 3 ×10⁸ ) / 656 × 10⁻⁹
= 3 × 10⁻¹⁹ J = 1.9 eV
En = - 13.6 / n² eV
where n = 1,2,3 ....
1) n = 2 to n = 1
E2 - E1 = - 13.6 eV ( 1 / 4 - 1/ 1)
= 10.2 eV
2) n = 3 to n = 2
E3 - E2 = -13.6 eV ( 1/3² - 1/ 2² )
= 1.9 eV
Thus the initial and the final states are n = 3 to n = 2.
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