Home
Class 12
PHYSICS
The wavelength of the first line of Balm...

The wavelength of the first line of Balmer series is `6563 Å`. The Rydberg's constant is

A

`1.09xx10^(5) m^(-1)`

B

`1.09xx10^(6) m^(-1)`

C

`1.09xx10^(7) m^(-1)`

D

`1.09xx10^(8) m^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
For longest wavelength of Balmer series,
`p=2` and `n=3`
`1/lambda=R(1/p^(2)-1/n^(2))=R(1/4-1/9)=(5R)/36`
`:. R=36/(5 lambda)=36/(5xx6.563xx10^(-7))`
`=1.09xx10^(7) m^(-1)`
Promotional Banner

Topper's Solved these Questions

  • CIRCULAR MOTION

    NIKITA PUBLICATION|Exercise Multiple Choice Question|421 Videos

Similar Questions

Explore conceptually related problems

The wavelength of the first line of Balmer series is 6563 Å . The Rydbergs constant fro hydrogen is about

The wavelength of H_(alpha) line in Balmer series is

The wavelength of first line of Balmer series is 6563Å . The wavelength of first line of Lyman series will be

the wavelength of the first line of lyman series is 1215 Å , the wavelength of first line of balmer series will be

If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å , the wavelngth of the second line of the series should be

The wavelength of the first line of Balmer series in hydrogen atom is 6562.8Å . Calculate ionisation potential of hydrogen and also, the wavelength of first line of Lyman series.

Assertion(A): Wavelength of limiting line of lyman series is less less than wavelength of limiting line of Balmer series. Reason(R): Rydberg constant value is same for all elements

The wavelength of the third line of the Balmer series for a hydrogen atom is -