The first member of the Balmer series of hydrogen atom has a wavelength of 6561 A ∘ . The wavelength of the second member of the Balmer series (in nm) is_____________

1 λ = R Z 2 1 n 1 2 - 1 n 2 2 1 λ 1 = R 1 2 1 2 2 - 1 4 2 = 3 R 16 λ 2 λ 1 = 20 27 λ 2 = 20 27 × 6561 A ∘ = 4860 A ∘ = 486 nm

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PhysicsModern PhysicsAtomic structureMedium2 minPYQ_2020

PhysicsMediumnumerical

The first member of the Balmer series of hydrogen atom has a wavelength of $6561 \overset{\circ}{A}$ . The wavelength of the second member of the Balmer series (in nm) is_____________

Answer:

486.00

Solution:

$\frac{1}{λ} = R Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$\frac{1}{λ_{1}} = R {(1)}^{2} (\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3 R}{16}$
$\frac{λ_{2}}{λ_{1}} = \frac{20}{27}$
$λ_{2} = \frac{20}{27} \times 6561 \overset{\circ}{A} = 4860 \overset{\circ}{A}$
$= 486 nm$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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