Atomic StructureHard
Question
The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is (ao is the Bohr radius)
a) $\frac{h^{2}}{4\pi^{2}ma_{o}^{2}}$ (b) $\frac{h^{2}}{16\pi^{2}ma_{o}^{2}}$ (c) $\frac{h^{2}}{32\pi^{2}ma_{o}^{2}}$ (d) $\frac{h^{2}}{64\pi^{2}ma_{o}^{2}}$
Solution
$K.E. = \frac{1}{2}mv^{2} = \frac{1}{2}m.\left( \frac{nh}{2\pi.mr} \right)^{2} = \frac{n^{2}h^{2}}{8\pi^{2}mr^{2}} = \frac{n^{2}h^{2}}{8\pi^{2}m.\left( a_{0}^{2}.n^{4} \right)} = \frac{h^{2}}{8\pi^{2}ma_{0}^{2}.n^{2}}$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
When β-particles are sent through a tin metal foil, most of them go straight through the foil as...Cathode rays are made up of electrons. Anode rays are made up of...The triad of nuclei that is isotonic is...Which of the the following has maximum energy ?...′e′ configuration of element with minimum size will be :-...