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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2019

PhysicsEasysingle choice

The wavelength of the ﬁrst line in Balmer series in the hydrogen spectrum is ‘ $λ$ ’. What is the wavelength of the second line in the same series?

Options:

Answer:

Solution:

For Balmer series.
$\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})$
$\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{9})$
$\frac{1}{λ} = \frac{5 R}{36}$ …(1)
For second line
$n = 4$
$\frac{1}{λ^{1}} = R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R}{16}$ …(2)
By equation (1) $\div$ (2)
$\frac{λ^{1}}{λ} = \frac{5}{38} \times \frac{16}{3}$
$\Rightarrow λ^{1} = \frac{20}{27} λ$

Stream:MHT_CETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2019

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