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  • For the Balmer series in the spectrum of H atom, <img alt="\bar{v}=\mathrm{R}_{\mathrm{H}}\left\{\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right\}" src="http://entrancecorner.oncodecogs.com/gif.latex?%5Cbar%7Bv%7D%3D%5Cmathrm%7BR%7D_%7B

For the Balmer series in the spectrum of H atom, \bar{v}=\mathrm{R}_{\mathrm{H}}\left\{\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right\} the correct statements among (I) to (IV) are: (1) As wavelength decreases, the lines in the series converge
(II) The integer n1 is equal to 2
(III) The lines of longest wavelength corresponds to n2 = 3
(IV) The ionization energy of hydrogen can be calculated from wavenumber of these lines
 
Option: 1 (II),(III),(IV)
Option: 2 (I),(III),(IV)
Option: 3 (I),(II),(III)
Option: 4 (I),(II),(IV)
 

Answers (1)

best_answer

Line Spectrum of Hydrogen-like atoms

\frac{1}{\lambda }= RZ^{2}\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )

Where R is called Rydberg constant, R = 1.097 X 107,  Z is the atomic number

n1= 1, 2, 3….

n2= n1+1, n1+2 ……

Lyman Series spectrum:

Where

n1= 1 and  n2= 2, 3, 4....

This lies in the Ultraviolet region.

Balmer Series Spectrum:

Where n1= 2 and  n2= 3, 4, 5, 6....

It lies in the visible region.

Paschen, Bracket and Pfund Series spectrums:

these lies in Infrared Region.

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Since, \mathrm{\Delta E\: \propto\:\frac{1}{\lambda} } so for wavelength to be longest or maximum the energy gap should be minimum. For the Balmer series, the transition for the longest wavelength is from n = 3 to n = 2. Thus clearly, n1 value for the Balmer series is 2. Further, as wavelength decreases, the lines in the series converge.

Therefore, Option(3) is correct.

Posted by

vishal kumar

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