Question

The electron in a hydrogen atom falls from an excited energy level to the ground state in two steps, causing the emission of photons with wavelengths of 1870 nm and 102.5 nm, respectively. what is the principal quantum number or shell of the initial excited energy level from which the electron falls?

Answer 1

Answer:

n = 4

Explanation:

As we know that the energy of the photon for a given wavelength is given as

$E = \frac{hc}{\lambda}$

so it is given as

$E_1 = \frac{1240}{1870}$

$E_1 = 0.66 eV$

similarly for other photon we have

$E_2 = \frac{1240}{102.5}$

$E_2 = 12.09 eV$

So we have

$E = E_1 + E_2$

$E = 12.09 + 0.66$

$E = 12.75$

so we have

$E = \frac{13.6}{n^2}$

$12.75 = 13.6(1 - \frac{1}{n^2})$

$n = 4$