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A photon of wavelength 300 nm interract...

A photon of wavelength 300 nm interracts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon I transferred to the electron os the atom. State which possibility is correct. (Consider,Planck's constant `=4xx10^(-15)eV *s , ` velocity of light `=3xx10^(8)m//s` , ionisation energy of hydrogen =13.6 eV.

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

Text Solution

Verified by Experts

The correct Answer is:
D
Energy of the photon
`E=hf=(hc)/(lambda)=((4xx10^(-15))xx(3xx10^(8)))/(300xx10^(-9))=4 eV`
Ground state energy of hydrogen atom =-13.6 eV
Energy of the second orbit =`-(13.6)/(2^2)=-3.4 =eV`
`therefore` Energy required to move the electron from ground state to second orbit =-3.4 -(-13.6)=10.2 eV.
Hence , the electron will keep orbiting in the ground state of atom .
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