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 the value of n is (hc = 1242 eV nm)

Answer 1

E_photon = E_{ionize atom} + E_{kinetic energy}

\(\frac{1242}{90}=\frac{13.6}{n^2}+10.4\)

from this, n = 2