The angular momentum of an electron in the hydrogen atom is 2 π 3 h , where h is plank's constant. The kinetic energy of this electron is:

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The angular momentum of an electron in the hydrogen atom is $\frac{3 h}{2 π}$ , where h is plank's constant. The kinetic energy of this electron is:

Answer:

Solution:

The angular momentum $L$ is given by $L = n ℏ = \frac{nh}{2 π}$ , where $n$ is the principal quantum number.

Given $L = \frac{3 h}{2 π}$ , we have $n = 3$ .

The energy of an electron in the $n$ th orbit of a hydrogen atom is $E_{n} = - \frac{13.6}{n ^{2}}$ eV.

For $n = 3$ , $E_{3} = - \frac{13.6}{3 ^{2}} = - \frac{13.6}{9} \approx - 1.51$ eV.

The kinetic energy is the magnitude of the total energy, so $K E = ∣ E_{3} ∣ = 1.51$ eV.

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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