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 atom

Answer

D

Explanation

`E_("ph) = (hc)/lamda(eV) = (4 xx10^(-15)xx3xx10^8)/(300 xx10^(-9))=4eV`
`1^(st)` excitation energy = 10.2eV for an electron
Since, `E_(ph) lt ` 10.2eV, no excitation in possible.