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PhysicsModern PhysicsAtomic structureEasy2 minPYQ_2025

PhysicsEasynumerical

An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{th}$ energy state by emitting a photon of energy 2.86 eV. The integer value of $n$ will be ____.

Answer:

2.00

Solution:

$E = 13.6 (\frac{1}{n _{1}^{2}} - \frac{1}{n _{1}^{2}})$
$2.86 = 13.6 (\frac{1}{n ^{2}} - \frac{1}{5 ^{2}})$
$\frac{1}{n ^{2}} = 0.21 + \frac{1}{2.5} n^{2} = 4 n = 2$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2025

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