The shortest wavelength of hydrogen atom in Lyman series is . The longest wavelength in Balmer series of is [2023]

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Solution

Q.
The shortest wavelength of hydrogen atom in Lyman series is $λ$ . The longest wavelength in Balmer series of ${He}^{+}$ is [2023]

1 $\frac{9 λ}{5}$

2 $\frac{36 λ}{5}$

3 $\frac{5 λ}{9}$

4 $\frac{5}{9 λ}$

Ans.
(1)

$(1) For Hydrogen atom (Z = 1) :$

${(\frac{1}{λ})}_{H} = R_{H} {(1)}^{2} [\frac{1}{1^{2}} - \frac{1}{\infty^{2}}] \to Last line$

$(2) {For He}^{+} atom (Z = 2) :$

${(\frac{1}{λ})}_{{He}^{+}} = R_{H} {(2)}^{2} [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = R_{H} [1 - \frac{4}{9}]$

$= R_{H} (\frac{5}{9}) \to first line$

$From (1) and (2)$

$\frac{λ_{{He}^{+}}}{λ_{H}} = \frac{9}{5}$

$λ_{{He}^{+}} = \frac{9 λ_{H}}{5}$

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