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Balmer given an equation for wavelength of visible radiation of H-spectrum as `lambda=(kn^(2))/(n^(2)-4)`.The value of k in terms of Rydbrum constant R is

A

`4 R`

B

`R//4`

C

`4//R`

D

`R`

Text Solution

Verified by Experts

The correct Answer is:
C

`1/(lambda) = R(1/4 - 1/(n^2))`
`= R((n^(2)-4)/(4n^2))`.
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