A photon of wavelength 300nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon is transferred to the electron of the atom. State which possibility is correct, (consider, Plank's constant `=4xx10^(-15)` eVs, velocity of light `=3xx10^(8)ms^(-1)` ionization energy of hydrogen =13.6 eV)

Options

A. Electron will be knocked out of the atom
B. Electron will go to any excited state of the atom
C. Electron will go only to first excited state of the atom
D. Electron will keep orbiting in the ground state of the atom

Answer

D

Explanation

The energy of the photon,
`" " E=(hc)/(lambda)=(4xx10^(-15)eVsxx3xx10^(8)ms^(-1))/(300xx10^(-9)m)=4eV`
The ionisation energy is 13.6eV which is greater than energy of photon, so than cannot come into excited state and will ramain in ground state.