Question

Derive the expression for frequency of radiation emitted by an electron transition

from higher energy level to lower energy level. ​

Answer 1

Explanation:

Using Bohr’s postulates, derive the expression for the frequency of radiation emitted when electron in hydrogen atom undergoes transition from higher energy state (quantum number ni) to the lower state, (nf).

When electron in hydrogen atom jumps from energy state ni =4 to nf =3, 2,1, identify the spectral series to which the emission lines belong.

Answer 2

To derive:

Expression for frequency of radiation emitted by an electron transition from higher energy level to lower energy level.

Derivation:

We know that transition of electron with difference in energy level orbits will result in formation of an electromagnetic wave with the same energy content.

Energy of electron in each orbit is given as:

$\therefore \: E = - \dfrac{13.6}{ {n}^{2} } \: eV$

So, energy difference is :

$\therefore \: \Delta E = E_{2}-E_{1}$

$\implies \: \Delta E = 13.6 \bigg \{ \dfrac{1}{ {(n_{2}) }^{2} } - \dfrac{1}{ {(n_{1}) }^{2} } \bigg \}$

Now, Energy can be written in terms of frequency as :

$\boxed{ \therefore\: E = h \nu}$

Continuing with previous equation:

$\implies \: h \nu= 13.6 \bigg \{ \dfrac{1}{ {(n_{2}) }^{2} } - \dfrac{1}{ {(n_{1}) }^{2} } \bigg \}$

$\implies \: \nu= \dfrac{13.6}{h} \bigg \{ \dfrac{1}{ {(n_{2}) }^{2} } - \dfrac{1}{ {(n_{1}) }^{2} } \bigg \}$

So, the final answer is:

$\boxed{ \bf\: \nu= \dfrac{13.6}{h} \bigg \{ \dfrac{1}{ {(n_{2}) }^{2} } - \dfrac{1}{ {(n_{1}) }^{2} } \bigg \}}$