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A difference of 2.3 eV separates two ene...

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits form the upper level to the lower level.

A

`6.65xx10^(14)Hz`

B

`3.68xx10^(15)Hz`

C

`5.5xx10^(14)Hz`

D

`9.11xx10^(15)Hz`

Text Solution

Verified by Experts

The correct Answer is:
C
Given, `E_(2)=E_(1)=2.3eV`
or `v=(E_(2)-E_(1))/(h)=(2.3xx1.6xx10^(-19))/(6.6xx10^(-34))`
`=0.55xx10^(15)`
`=5.5xx10^(14)Hz`
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