Question

One of the lines in the Balmer series of the hydrogen atom emission spectrum is at 397 nm. It results from a transition from an upper energy level to n = 2. What is the principal quantum number of the upper level?

Answer 1

Given n = 2, λ = 397 nm.

We have to find Principal quantum number of the upper level. That is $n^{ \prime }$

Change in the energy is given by

$\frac { 1 }{ \lambda } =\quad { R }_{ h }\left\{ \frac { 1 }{ { n }^{ 2 } } -\frac { 1 }{ { n\prime }^{ 2 }} \right\} where\quad{ R }_{ h }= Rydberg constant.$

Substituting the values of n and λ,  

$\frac { 1 }{ 397\times { 10 }^{ -9 } } =1.097\times { 10 }^{ 7 }\left\{ \frac { 1 }{ { 2 }^{ 2 } } +\frac { 1 }{ { n\prime }^{ 2 } }\right\}$

$\left\{ \frac { 1 }{ 4 } -\frac { 1 }{ { n\prime }^{ 2 } }\right\} =\quad 0.2296\\ \\ \frac { 1 }{ { n\prime }^{ 2 } } =0.25-0.2296\quad =0.00204\\\\ { n\prime}^{ 2 }=\frac { 1 }{ 0.00204 } =49\\\\ { n\prime }=7$

Therefore, Principal quantum number = 7.

Answer 2

HEY MATE....

PRINCIPAL QUANTUM NO. = 7

I HOPE IT IS HELPFUL.....☺✌☺