Question #118771
The bound electron in a hydrogen atom starts off in an excited state and releases a photon with a wavelength of 95.0 nm as it transitions to the ground state.

What orbital (n) did the electron start in?
1
Expert's answer
2020-05-29T09:59:18-0400

Wavelength of photon is λ=95nm\lambda=95nm .

Let us consider the electron was in n=n1n=n_1 orbital and transits to n=1n=1.


Now, from Bohr's principle of hydrogen atom,we know that the amount of energy released by electron in transition is equal to energy of the emitted photon, thus we get

Δ=hcλ=13.6(1121n12)eV    1240eVnmλ=13.6(1121n12)eV    1240eVnm95nm=13.6(1121n12)eV    1n12=10.9597=0.0403    n1=10.0403=4.98\Delta=\frac{hc}{\lambda}=13.6\bigg(\frac{1}{1^2}-\frac{1}{n_1^2}\bigg)eV\\ \implies \frac{1240eV\cdot nm}{\lambda}=13.6\bigg(\frac{1}{1^2}-\frac{1}{n_1^2}\bigg)eV\\ \implies \frac{1240eV\cdot nm}{95nm}=13.6\bigg(\frac{1}{1^2}-\frac{1}{n_1^2}\bigg)eV\\ \implies \frac{1}{n_1^2}=1-0.9597=0.0403‬\\\implies n_1=\sqrt{\frac{1}{0.0403‬}}=4.98

But,nn is always integer hence,n1=5n_1=5


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