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Physics - Modern Physics Question with Solution | TestHub

PhysicsModern PhysicsAtomic structureHard2 minQB

PhysicsHardsingle choice

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

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Answer:

Solution:

$\frac{1}{λ} = Rz^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}}) \frac{\frac{1}{λ _{L}} = Rz ^{2} ( \frac{1}{1 ^{2}} )}{\frac{1}{λ _{B}} = Rz ^{2} ( \frac{1}{2 ^{2}} )} \frac{λ _{B}}{λ _{L}} = 4 : 1$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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