Answer:
D. 160 nm
Explanation:
The energy released from n = 3 to n = 1 must be equal to the sum of energies released from n = 3 to n = 2 and from n = 2 to n = 1. Therefore,
Energy of Photon from 3 to 1 = Energy of Photon from 3 to 2 + Energy of Photon from 2 to 1
[tex]\frac{hc}{\lambda} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}\\\\\frac{1}{\lambda} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}}[/tex]
where,
λ = wavelength of photon released from 3 to 1 = ?
λ₁ = wavelength of photon released from 3 to 2 = 800 nm
λ₂ = wavelength of photon released from 2 to 1 = 200 nm
Therefore,
[tex]\frac{1}{\lambda} = \frac{1}{800\ nm} + \frac{1}{200\ nm}\\\\\frac{1}{\lambda} = 0.00625 nm^{-1}\\\\\lambda = \frac{1}{0.00625 nm^{-1}}\\\\\lambda = 160 nm[/tex]
Therefore, the correct option is:
D. 160 nm