Question

Consider a hydrogen atom with its electron in the $n^th$ orbital. An electromagnetic radiation of wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then value of n is (hc = 1242 eV nm)

Answer 1

Energy of photon $E_{Ph} = \frac{1242}{\lambda\left(in \,nm\right)} eV$
$λ = 90\, nm$
$E_{Ph} = \frac{1242}{90}eV ⇒ E_{Ph} = 13.8 eV$
K.E. of ejected electron K.E. = 10.4 eV
So let energy of the orbits is $E_{n}$
$E_{n} + E_{Ph} = 10.4 eV$
$E_{n} + 13.8 = 10.4$
$E_{n} = −3.4 eV$
$E_{n} = - 13.6 \frac{Z^{2}}{n^{2}}eV$
for H-atom $\left(Z = 1\right)$
$- 3.4 = - 13.6 \frac{\left(1\right)^{2}}{n^{2}}$
$n = 2$