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An electron jumps from the 4th orbit to ...

An electron jumps from the `4th` orbit to the `2nd` orbit of hydrogen atom. Given the Rydberg's constant `R = 10^(5) cm^(-1)`. The frequency in `Hz` of the emitted radiation will be

A

`3/16 xx 10^5`

B

`3/16 xx 10^(15)`

C

`9/16 xx 10^(15)`

D

`3/4 xx 10^(15)`

Text Solution

Verified by Experts

The correct Answer is:
C
`v=(c )/( lamda) = cR ((1)/( p^2)-(1)/(n^2)) = cR ((1)/(4) -(1)/(16))`
`= ( 3 xx 10^8 xx 10 ^7 xx 12 ) /( 64 ) = ( 9 )/( 16 ) xx 10^(15)` Hz
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DISHA PUBLICATION-ATOMS-EXERCISE-1: CONCEPT BUILDER (TOPICWISE)
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