The angular momentum of an electron in hydrogen atom is h/(π). The kinetic energy of the electron is

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Correct answer: (B) 3.4 eV. The angular momentum of an electron in hydrogen atom =h/(π) . As we know, from Bohr's second postulate, mvr=h/(π)⇒v=h/(πmr) ...(i) Now, kinetic energy of electron =1/2mv^2=1/2m×h^(πmr2) [from Equation (i)] =(h^2)/(2π^2mr^2) =(6.63×10^(-34))/(22×3.14^2×9.1×10^(-31)×5.3×10^(-11)2×1.6×10^(-19)) =3.4 eV

The angular momentum of an electron in hydrogen atom is $\frac{h}{π}$ . The kinetic energy of the electron is

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More Atomic Physics Questions

The angular momentum of an electron in hydrogen atom is hπ. The kinetic energy of the electron is

Solution

More Atomic Physics Questions

The angular momentum of an electron in hydrogen atom is $\frac{h}{π}$ . The kinetic energy of the electron is