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## Q. 1.2

Find the wavelength of the photon emitted when a hydrogen atom goes from n = 10 state to the ground state.

## Verified Solution

The wavelength in Angstrom units is given by,$\lambda =\frac{12,400}{E_{2}-E_{1} }$

Since the hydrogen atom goes from n = 10 state to the ground state,$\lambda =\frac{12,400}{E_{10}-E_{1} }$

The energy of the $10^{th}$ state is $E_{10}=\frac{-13.6}{10^{2} }=-0.136eV.$

The energy in the ground state is $E_{1}=-13.6eV.$

Thus the wavelength of the emitted photon $=\frac{12,400}{-0.136-(-13.6)}=920.97$ Å