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For the hydrogen atom the energy of rad...

For the hydrogen atom the energy of radiation emitted in the transitation from 4th excited state
to 2nd exicited state according to Bohr 's theory is

A

0.57 eV

B

0.667 eV

C

0.967 eV

D

1.267 eV

Text Solution

Verified by Experts

The correct Answer is:
C
Using `E_(n)=(13.6)/(n^(2))` eV
according to given question
`E_(4th)=-(13.6)/(4)^(2)=-(13.6)/(16)eV=-0.85 eV`
energy of radiation emitted in the transition
`Delta E= E_(4th)-E_(2nd)=--0.85+3.4 =2.55 eV`
`Delta E= E_(3rd)-E_(2nd)=-1.51 +3.41=1.89 eV`
`DeltaE= E_(4th)-e_(3rd)=-0.85+1.51 =0 0.66 eV`
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