Question

What wavelength photon is required to excite a hydrogen atom from the n = 1 state to the n = 3 state?

Answer 1

The wavelength required is 102.9 nm.

Step-by-step explanation:

The energy levels for the hydrogen atom are

`E(n) = (-13.6 \ eV)/(n^2)`

So, for a transition from the first level to the third level we got

`\Delta E = E(3) - E (1)`

`\Delta E = (-13.6 \ eV)/( 3 ^2) - (-13.6 \ eV)/(1^2)`

`\Delta E = (-13.6 \ eV)/( 9) - (-13.6 \ eV)/(1)`

`\Delta E = (8)/(9) 13.6 \ eV`

`\Delta E = 12.09 \ eV`

`\Delta E = 12.09 \ eV * (1.6 \ 10^-19 \ Joules)/(1 \ eV)`

`\Delta E = 1.93 \ 10^-18 \ Joules`

So we need a photon with this energy.

The energy of a photon its given by

`E = h \\u = h (c)/(\lambda)`

So, the wavelength will be

`\lambda = (h c)/(E)`

`\lambda = (6.62 \ 10^(-34) \ (m^2 kg)/(s) \ *  3.00 \ 10^8 \ (m)/(s))/(1.93 \ 10^-18 \ Joules)`

`\lambda = 10.29 \ 10^(-8) m`

`\lambda = 1.029 \ 10^(-7) m`

`\lambda = 102.9 \ nm`