The spectral line of the shortest wavelength in Balmer series of atomic hydrogen will be:

6.215 \times 10^{- 5} cm

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1.437 \times 10^{- 5} cm

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5.942 \times 10^{- 5} cm

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3.647 \times 10^{- 5} cm

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Solution

The correct option is D $3.647 \times 10^{- 5} cm$
Rydberg formula for hydrogen atom is given by
$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
For Balmer series $n_{1} = 2$ and for minimum wavelength $n_{2} = \infty$
Thus the shortest wavelength will be
$\frac{1}{λ_{s h o r t e s t}} = 109678 \times 1^{2} \times (\frac{1}{2^{2}} - \frac{1}{\infty^{2}}) λ_{s h o r t e s t} = 3.647 \times 10^{- 5} cm$

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Emission and Absorption Spectra

Standard XII Chemistry

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The spectral line of the shortest wavelength in Balmer series of atomic hydrogen will be:

The correct option is D 3.647×10−5 cmRydberg formula for hydrogen atom is given by 1λ=RZ2(1n21−1n22) For Balmer series n1=2 and for minimum wavelength n2=∞ Thus the shortest wavelength will be 1λshortest=109678×12×(122−1∞2)λshortest=3.647×10−5 cm