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  • Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from <img alt="\mathrm{n}=4" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from \mathrm{n}=4 to \mathrm{n}=2 of \mathrm{He}^{+}spectrum

Option: 1

\mathrm{n}=2 to \mathrm{n}=1


Option: 2

\mathrm{n}=1 to \mathrm{n}=2


Option: 3

\mathrm{n}=3 to \mathrm{n}=4


Option: 4

\mathrm{n}=1 to \mathrm{n}=3


Answers (1)

best_answer

\lambda_{\mathrm{H}}=\lambda_{\mathrm{He}^{+}}

\mathrm{R}_{\mathrm{H}} \times(1)^{2}\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)=\mathrm{R}_{\mathrm{H}} \times(2)^{2}\left(\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right)
\left(\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right)=\left(\frac{4}{4}\right)-\left(\frac{4}{16}\right)

\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}=\frac{1}{1}-\frac{1}{4}
\mathrm{n}_{1}=1: \mathrm{n}_{2}=2$ for $\mathrm{H} -atom

Posted by

HARSH KANKARIA

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