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)` ionisation 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

Here, wavelength of photon, `lambda=300 nm`
`h=4 times 10^(-15)eVs, c=3 times 10^(8)m" "s^(-1)`
Ionization energy of hydrogen = 13.6 eV
Energy of photon, `E=(hc)/(lambda)`
`" "=((4 times 10^(-15)eVs)(3 times 10^(8)ms^(-1)))/((300 times 10^(-9)m))=4eV`
For hydrogen atom, `E_(n)=-(13.6)/n^(2)eV`
First excitation energy `=E_(2)-E_(1)`
`" "=-13.6/2^(2)-(-13.6/1^(2))=-3.4+13.6=10.2eV`
Since energy of photon is less than 10.2 eV, so no excitation is possible, i.e., electron will keep orbiting in the ground state of atom.