The angular momentum of an electron in the hydrogen atom is 3 h-2 - Here h is planck-s constant- The kinetic energy of this electron isA- 4-53 eVB- 3-4 eVC- 1-51 eVD- 6-8 eV

The correct option is C 1.51 eVAngular momentum, L = nh2 π = 3h2 π (given) also n = 3 Kinetic energy, KE = total energy ∴ KE = 13.6n^2 eV KE = 13.6(3)^2 eV KE ...

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Standard XII

Physics

Angular Analogue of Linear Momentum

The angular m...

Question

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

4.53 eV

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1.51 eV

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3.4 eV

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6.8 eV

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Solution

The correct option is B $1.51 eV$
Angular momentum,
$L = \frac{n h}{2 π} = \frac{3 h}{2 π}$ (given)
also $n = 3$
$Kinetic energy, KE = - total energy$
$∴ K E = \frac{13.6}{n^{2}} eV$
$K E = \frac{13.6}{(3)^{2}} eV$
$K E = \frac{13.6}{9} eV$
$K E = 1.51 eV$

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The angular momentum of an electron in the hydrogen atom is 3h2π. Here h is planck's constant. The kinetic energy of this electron is

The correct option is B 1.51 eVAngular momentum, L=nh2π=3h2π (given) also n=3 Kinetic energy, KE=−total energy ∴KE=13.6n2 eV KE=13.6(3)2 eV KE=13.69 eV KE=1.51 eV

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

The correct option is B $1.51 eV$
Angular momentum,
$L = \frac{n h}{2 π} = \frac{3 h}{2 π}$ (given)
also $n = 3$
$Kinetic energy, KE = - total energy$
$∴ K E = \frac{13.6}{n^{2}} eV$
$K E = \frac{13.6}{(3)^{2}} eV$
$K E = \frac{13.6}{9} eV$
$K E = 1.51 eV$