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_____________

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

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PhysicsModern PhysicsAtomic structureMedium2 minQB

PhysicsMediumnumerical

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

Answer:

486.00

Solution:

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

$\frac{1}{λ _{1}} = R (1)^{2} (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}}) = \frac{3 R}{16}$

$\frac{λ _{2}}{λ _{1}} = \frac{20}{27}$

$λ_{2} = \frac{20}{27} \times 6561 A \circ = 4860 A \circ$

$= 486 nm$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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