Question

What is the wavelength and frequency of a photon emitted by transition of an electron from a n- orbit to a n-1 orbit'?

Answer 1

`\lambda=9.12* 10^(-8)}* \frac {{{{(n-1)}^2}* n^2}}{1-2n}\ m`

`\\u=3.29* 10^(15)\frac{1-2n}{{{(n-1)}^2}* n^2}}\ s^(-1)`

Step-by-step explanation:

`E_n=-2.179* 10^(-18)* (1)/(n^2)\ Joules`

For transitions:

`Energy\ Difference,\ \Delta E= E_f-E_i =-2.179* 10^(-18)((1)/(n_f^2)-(1)/(n_i^2))\ J=2.179* 10^(-18)((1)/(n_i^2) - (1)/(n_f^2))\ J`

`n_i=n\ and\ n_f=n-1`

Thus solving it, we get:

`\Delta E=2.179* 10^(-18)((1)/(n^2) - \frac{1}{{(n-1)}^2})\ J`

`\Delta E=2.179* 10^(-18)(\frac{{(n-1)}^2-n^2}{{{(n-1)}^2}* n^2}})\ J`

`\Delta E=2.179* 10^(-18)(\frac{n^2+1-2n-n^2}{{{(n-1)}^2}* n^2}})\ J`

`\Delta E=2.179* 10^(-18)(\frac{1-2n}{{{(n-1)}^2}* n^2}})\ J`

Also,
`\Delta E=\frac {h* c}{\lambda}`

Where,

h is Plank's constant having value
`6.626* 10^(-34)\ Js`

c is the speed of light having value
`3* 10^8\ m/s`

So,

`\frac {h* c}{\lambda}=2.179* 10^(-18)(\frac{1-2n}{{{(n-1)}^2}* n^2}})\ J`

`\lambda=\frac {6.626* 10^(-34)* 3* 10^8}{2.179* 10^(-18)}* \frac {{{{(n-1)}^2}* n^2}}{{1-2n}}\ m`

So,

`\lambda=9.12* 10^(-8)}* \frac {{{{(n-1)}^2}* n^2}}{1-2n}\ m`

Also,
`\Delta E=h* \\u`

So,

`h* \\u=2.179* 10^(-18)\frac{1-2n}{{{(n-1)}^2}* n^2}}`

`\\u=\frac {2.179* 10^(-18)}{6.626* 10^(-34)}\frac{1-2n}{{{(n-1)}^2}* n^2}}\ s^(-1)`

`\\u=3.29* 10^(15)\frac{1-2n}{{{(n-1)}^2}* n^2}}\ s^(-1)`