JEE Mains · Chemistry · STD 11 - 2. structure of atom
For the Balmer series in the spectrum of \(\mathrm{H}\) atom, \(\quad \bar{v}=R_{H}\left\{\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right\}, \quad\) the correct statements among \((I)\) to \((IV)\) are \((I)\) As wavelength decreases, the lines in the series converge \((II)\) The integer \(n_{1}\) is equal to \(2\) \((III)\) The lines of longest wavelength corresponds to \(\mathrm{n}_{2}=3\) \((IV)\) The ionization energy of hydrogen can be calculated from wave number of these lines
- A \((II), (III), (IV)\)
- B \((I), (II), (III)\)
- C \((I), (III), (IV)\)
- D \((I), (II), (IV)\)
Answer & Solution
Correct Answer
(B) \((I), (II), (III)\)
Step-by-step Solution
Detailed explanation
For balmer : \(\mathrm{n}_{1}=2, \mathrm{n}_{2}=3,4,5, \ldots \infty\) \(\bar{v}=\frac{1}{\lambda}=R_{H}\left[\frac{1}{2^{2}}-\frac{1}{n_{2}^{2}}\right]\) \(\frac{1}{\lambda_{\text {longest }}}=\mathrm{R}_{\mathrm{H}}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]\)
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