Question

Find the number of spectral lines in the Paschen series emitted by atomic $ H $ ,when electrons excited from ground state to $ {70^{th}} $ energy level return back.

Answer 1

Hint: We know that in the Lyman series the electron comes back to the ground state that is the first energy level. In the Balmer series the electron returns back to the second energy level . In the Paschen series the electron returns back to the third energy level. When there is a transition of an electron between two energy levels it forms spectral lines.

Complete step by step answer
We know that in the Paschen series the transition is between the third energy level to the energy level greater than three. We will denote the energy level with $ {n_i} $ where $ i = 1,2,3,4..... $ . . The $ {70^{th}} $ energy level can be written as $ {n_{70}} $ . The third energy level can be written as $ {n_3} $ .The number of spectral lines between two energy levels is given by the formula $ \dfrac{{(\Delta n)(\Delta n + 1)}}{2} $ . Here $ \Delta n $ is a change in energy level . In the question it is given that the electron is returning back from the $ {70^{th}} $ energy level to the ground state and we have to find the value of the number of spectral lines in the Paschen series. So we have to find the number of spectral lines in between $ {n_3} $ and $ {n_{70}} $ . The value of $ \Delta n $ will be $ \Rightarrow 70 - 3 = 67 $ . So, the number of spectral line in the Paschen series will be $ $ $ \Rightarrow \dfrac{{(\Delta n)(\Delta n + 1)}}{2} = \dfrac{{(67)(67 + 1)}}{2} = 2278 $ . So, from the above explanation and calculation it is clear to us that the correct answer of the given question is $ 2278 $ .

Note
Always remember that the number of spectral lines formed when there is a transition of an electron between two energy levels is given by the formula $ \dfrac{{(\Delta n)(\Delta n + 1)}}{2} $ .Here $ \Delta n $ is change in energy level. Always avoid silly mistakes and calculation errors.