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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

PhysicsEasysingle choice

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

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Solution:

$\frac{1}{λ_{1}} = R [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = \frac{5}{36} R$

$\frac{1}{λ_{2}} = R [\frac{1}{2^{2}} - \frac{1}{4^{2}}] = R [\frac{3}{16}]$

$\frac{λ_{2}}{λ_{1}} = \frac{5 R / 36}{3 R / 16} = \frac{5}{36} \times \frac{16}{3}$

$λ_{2} = \frac{5}{36} \times \frac{16}{3} \times 660 = 488.9 nm$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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