Home
Class 11
CHEMISTRY
The balmer series occurs between the wav...

The balmer series occurs between the wavelength of `[R = 1.0968 xx 10^7 m^-1]`.

Text Solution

Verified by Experts

For Balmer series,
`(1)/(lamda)=R[(1)/(2^(2))-(1)/(n^(2))]`
where, `n=3,4,5, . . Infty`
To obtain the limits for Balmer series n=3 and `n=infty` respectively.
`lamda_(max)(n=3)=(1)/(R[(1)/(2^(2))-(1)/(3^(2))])=(36)/(5R)`
`=(36)/(5xx1.0968xx10^(7))m=`6564Ã…
`lamda_("min")(n=infty)=(1)/(R[(1)/(2^(2))-(1)/(infty^(2))])=(4)/(R)`
`=(4)/(1.0968xx10^(7))m`
=3647Ã….
Promotional Banner

Topper's Solved these Questions

  • ATOMIC STRUCTURE

    OP TANDON|Exercise Illustrations of Objective Question|33 Videos

  • ATOMIC STRUCTURE

    OP TANDON|Exercise Solved Examples|10 Videos

  • AROMATIC HYDROCARBONS (ARENES)

    OP TANDON|Exercise EXAMPLES|24 Videos

  • CHARACTERISATION OF ORGANIC COMPOUNDS

    OP TANDON|Exercise Passage-2|5 Videos

Similar Questions

Explore conceptually related problems

Calculate the longest and shortest wavelength in the Balmer series of hydrogen atom. Given Rydberg constant = 1.0987xx10^7m^-1 .

The lines in Balmer series have their wavelengths lying between

Calculate longest wavelength of Paschen series. Given R=1.097xx10^(7)m^(-1) .

The wavelength of the first member of the balmer series of hydrogen is 6563xx10^(-10)m . Calculate the wavelength of its second member.

Calculate the wavelength and energy for radiation emitted for the electron transition from infinite (oo) to stationary state of the hydrogen atom R = 1.0967 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34) J s and c = 2.979 xx 10^(8) m s^(-1)

The element which has a K_(alpha) X-rays line of wavelength 1.8 Å is (R = 1.1 xx 10^(7) m^(-1), b = 1 and sqrt(5//33) = 0.39)