Question

The ground state energy of hydrogen atom is $-13.6eV$.
(ii) If the electron jumps to the ground state from the $2^{nd}$ excited state, calculate the wavelength of the spectral line emitted.

Answer 1

2nd excited state means 3rd normal state or n=3 state

total energy of $n^{th}level$ is $E_n=E_0/n^2$ ,

given that $E_0=-13.6eV$

So total energy for 3rd state will be $E_3=-13.6/9=-1.51eV$

Energy of the transition will be the difference in energy levels so $E=-1.51-(-13.6)=12.09eV$

Wavelength $\lambda =\frac{1237.5}{E\left(eV\right)}nm=\frac{1237.5}{12.09}nm=102.3nanometer$