Question

Calculate the energy of the third line in the brackett series for hydrogen atom?

Answer 1

The third line of Brackett series for hydrogen atom is  n' = 4 & n = 7.

$\text{We know that :}\\ \\ \frac{1}{\lambda} = Z^2R_H\Big(\frac{1}{n'^2}-\frac{1}{n^2}\Big) \\ \\ \implies \frac{1}{\lambda} = (1)^2 R_H\Big(\frac{1}{4^2}-\frac{1}{7^2}\Big) \\ \\ \implies \lambda = 2.2 \times 10^{-14} m \\ \\ \text{Also :} \\ \\ E= \frac{hc}{\lambda} \\ \\ \implies E=\frac{6.6\times 10^{-34}\times 3\times 10^8}{2.2\times 10^{-14}} \\ \\ E= 9\times 10^{-12} J$

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