Question

Calculate the wavelength, in nanometers, of the spectral line produced when an electron in a hydrogen atom undergoes the transition from the energy level ????=3n=3 to the level ????=1.

Answer 1

`\lambda=1026.28\ nm`

Step-by-step explanation:

The expression for energy is -

`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`

`\Delta E=2.179* 10^(-18)((1)/(n_i^2) - (1)/(n_f^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)(|(1)/(n_i^2) - (1)/(n_f^2)|)\ J`

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

So,

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

Given,
`n_i=3\ and\ n_f=1`

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

`\lambda=(10^(-26)* \:19.878)/(10^(-18)* \:2.179\left(|(1)/(9)-(1)/(1)\right)|)\ m`

`\lambda=(19.878)/(10^8* \:2.179\left(|-(8)/(9)\right|))\ m`

`\lambda=1.02628* 10^(-6)\ m`

1 m = 10⁻⁹ nm

`\lambda=1026.28\ nm`