Exercise 5 | Early Quantum Theory

Chapter 4 - Early Quantum Theory

What is the initial energy level when an electron of a hydrogen atom transitions into n = 2 and emits a photon at 410 nm?

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Solution

E_photon = $\frac{hc}{λ}$

and E_photon= $- 2.1799 \times 10^{- 18} \times (\frac{1}{{n_{i}}^{2}} - \frac{1}{{n_{f}}^{2}})$

⇒ $\frac{1}{{n_{i}}^{2}} - \frac{1}{4} = \frac{hc}{λ \times (- 2.1799 x 10^{- 18})}$ = -0.222

⇒ $\frac{1}{{n_{i}}^{2}}$ = 2.76 x 10^-2

⇒ n_i = 6