QJEEM20Ch0f1d: JEE Main 2020 Chemistry - Class 11 JEEM20Ch0f1d Solution: The shortest wavelength of H atom in the Lyman series is λ 1 .…

Question

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The shortest wavelength of H atom in the Lyman series is

λ

₁. The longest wavelength in the Balmar series of He⁺ is :

Accepted Answer

For Shortest wavelength energy should be maximum.

For maximum energy transition must be form n =

\infty

to n = 1.

\frac{1}{λ_{1}} = R_{H} {(1)}^{2} [\frac{1}{1} - 0]

\Rightarrow

\frac{1}{λ_{1}} = R_{H}

For longest wavelength,

Δ

E = minimum.
For Blamer series n = 3 to n = 2 will have

Δ

E minimum for He⁺, Z = 2

So

\frac{1}{λ_{2}} = R_{H} {(2)}^{2} [\frac{1}{4} - \frac{1}{9}]

\Rightarrow

\frac{1}{λ_{2}} = R_{H} \times \frac{5}{9}

\Rightarrow

λ_{2} = λ_{1} \times \frac{9}{5}

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