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In terms of Rydberg constant R, the shor...

In terms of Rydberg constant `R`, the shortest wavelength in the Balmer series of the hydrogen , atom spestrum will have wavelength

A

`1//R`

B

`4//R`

C

`3//2 R`

D

`9//R`

Text Solution

Verified by Experts

The correct Answer is:
B
Frequency `= R ((1)/(2^(2)) - (1)/(oo^(2))) C`
hence, `lambda = (4)/( R)`
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