Question

Consider a hydrogen atom with its electron in the $n^{t}h$ 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

Answer: 2

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