Question

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If the photon emitted has a wavelength of 487 nm, what is the value of ni?

Answer 1

Answer:

initial energy state is n = 4

Explanation:

For hydrogen atom when electron makes transition from higher energy state to lower energy state then the photon is released of energy which equal to the energy difference of two states

so here we can say

$h\nu = E_2 - E_1$

so here we have energy at nth excited state of electron is given as

$E_n = -13.6 \frac{z^2}{n^2} eV$

here z = 1 for hydrogen

so now we have

$\frac{hc}{\lambda} = -13.6 \frac{1^2}{n^2} + 13.6 \frac{1^2}{2^2}$

here we know that

$hc = 1242 eV-nm$

now plug in all values in it

$\frac{1242 eV-nm}{487 nm} = 13.6 eV(\frac{1}{4} - \frac{1}{n^2})$

$0.187 = 0.25 - \frac{1}{n^2}$

by solving above equation we got

n = 4