The time period of revolution of electron in its ground state orbit in a hydrogen atom is 1.6 × 10 - 16 s . The frequency of revolution of the electron in its first excited state (in s - 1 ) is:

T ∝ r v ∝ n 2 z × n z ∝ n 3 z 2 T 1 T 2 = n 1 3 n 2 3 = 1 8 T 2 = 8 T 1 now, f 2 = 1 T 2 f 2 = 1 12.8 × 10 - 16 ≈ 7.8 × 10 14

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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2020

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The time period of revolution of electron in its ground state orbit in a hydrogen atom is $1.6 \times 10^{- 16} s .$ The frequency of revolution of the electron in its first excited state (in $s^{- 1}$ ) is:

Options:

Answer:

Solution:

$T \propto \frac{r}{v} \propto \frac{n^{2}}{z} \times \frac{n}{z} \propto \frac{n^{3}}{z^{2}}$
$\frac{T_{1}}{T_{2}} = \frac{n_{1}^{3}}{n_{2}^{3}} = \frac{1}{8}$
$T_{2} = 8 T_{1}$

now,
$f_{2 = \frac{1}{T_{2}}}$
$f_{2} = \frac{1}{12.8 \times 10^{- 16}} \approx 7.8 \times 10^{14}$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2020

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