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PhysicsModern PhysicsAtomic structureHard2 minPYQ_2023

PhysicsHardnumerical

A light of energy $12.75 eV$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{π} \times 10^{- 17} eVs$ . The value of $x$ is ______ (use $h = 4.14 \times 10^{- 15} eVs, c = 3 \times 10^{8} m s^{- 1}$ )

Answer:

828.00

Solution:

Let the electron jumped to $n^{t h}$ excited state.

In the ground state, energy $E = - 13.6 eV$

So, using relation $12.75 = 13.6 [\frac{1}{1^{2}} - \frac{1}{n^{2}}]$

$\Rightarrow 0.9375 = [1 - \frac{1}{n^{2}}] \Rightarrow n^{2} = 16$

$\Rightarrow n = \sqrt{16} = 4$

Now, angular momentum, $L = \frac{n h}{2 π} = \frac{4 h}{2 π} = \frac{2 h}{π}$

$= \frac{2}{π} \times 4.14 \times 10^{- 15}$

$= \frac{828 \times 10^{- 17}}{π} eVs$

Hence, the value of $x = 828$ .

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: PYQ_2023

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