If the wavelength of 1 st line of Balmer series of hydrogen is 6561 A ˚ , the wavelength of the 2 nd line of series will be

For the first line of Balmer series, λ 1 1 = R ( 2 2 1 − 3 2 1 ) = 36 5 R For the second line of Balmer series. λ 2 1 = R ( 2 2 1 − 4 2 1 ) = 16 3 R ∴ λ 1 λ 2 = 3 R /16 5 R /36 = 27 20 ; or λ 2 = 27 20 ( 6561 A ˚ ) = 4860 A ˚

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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2014

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If the wavelength of $1^{st}$ line of Balmer series of hydrogen is $6561 \overset{˚}{A}$ , the wavelength of the $2^{nd}$ line of series will be

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For the first line of Balmer series, $\frac{1}{λ _{1}} = R (\frac{1}{2 ^{2}} - \frac{1}{3 ^{2}}) = \frac{5 R}{36}$ For the second line of Balmer series. $\frac{1}{λ _{2}} = R (\frac{1}{2 ^{2}} - \frac{1}{4 ^{2}}) = \frac{3 R}{16}$ $∴ \frac{λ _{2}}{λ _{1}} = \frac{5 R /36}{3 R /16} = \frac{20}{27}$ ; or $λ_{2} = \frac{20}{27} (6561 \overset{˚}{A}) = 4860 \overset{˚}{A}$

Stream:AIIMSSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2014

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