Question

An electron undergoes a transition from the n=3 state to the ground state in a hydrogen atom. Calculate the wavelength of the photon emitted as a result of this transition in the units of meters. An excellent response will include the appropriate relationship along with consideration of dimensional analysis , scientific notation, and consideration of significant figures.

Answer 1

`\lambda = 102.7 nm`

Step-by-step explanation:

As we know by Bohr's Postulates that when electron makes transition from higher excited state to lower excited state then energy difference of two levels is released in the form of photons.

So we can say

`(hc)/(\lambda) = E_2 - E_1`

here we know that energy of nth excited state in any atom is given as

`E_n = -13.6 (z^2)/(n^2) eV`

here for hydrogen atom z = 1

`n_2 = 3`

`n_1 = 1`

also we know that

`hc = 1242 eV -nm`

now from above equation we have

`(1242 eV-nm)/(\lambda) = (-13.6 (1^2)/(3^2))-(-13.6 (1^2)/(1^2))eV`

`(1242 eV-nm)/(\lambda) = 13.6 (1 - (1)/(9)) eV`

`\lambda = 102.7 nm`