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PhysicsModern PhysicsAtomic structureEasy2 minQB

PhysicsEasysingle choice

Taking the wavelength of first Balmer line in hydrogen spectrum $(n = 3$ to $n = 2)$ as $660 nm$ , the wavelength of the $2^{n d}$ Balmer line $(n = 4$ to $n = 2)$ will be :

Options:

Answer:

Solution:

From Rydberg’s equation,

$\frac{1}{λ} = R (\frac{1}{n ^{2}} - \frac{1}{n _{2}^{2}})$

$\frac{1}{660 \times 1 0 ^{- 9}} = R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}})$

$\frac{1}{λ} = R (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}})$

$\frac{λ}{600 \times 1 0 ^{- 9}} = \frac{5}{9 \times 4} \times \frac{16 \times 4}{12}$

$λ = 488.9 nm$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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