Question

Using Bohr's postulates for hydrogen atom,
show that the total energy (E) of the electron
in the stationary states can be expressed as
the sum of kinetic energy (K) and potential
energy (U), where K = -2U. Hence deduce
the expression for the total energy in the nth
energy level of hydrogen atom.​

Answer 1

Answer:

sorry for not knowing the ans

Answer 2

Answer:

Energy level of hydrogen atom.​$E_n=\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r}$

Explanation:

Step 1: According to Bohr's postulate, the force of gravitation between the nucleus and the electron provides the centripetal force to an electron rotating around the nucleus that is positively charged. The combination of kinetic and potential energy yields the total energy (T.E).

Bohr's hypotheses for the hydrogen atom: The hydrogen atom's electron can travel along any of the numerous circular paths with set radii and energies to circle the nucleus. Orbits, stationary states, and allowed energy states are the names of these routes.

Step 2: Bohr developed a three-postulate theory by fusing ideas from early quantum mechanics and classical mechanics. These suppositions are: 1. An atom's electron spins in a series of steady orbits without releasing any visible light.

The term "orbits" or "shells" refers to the idea that in an atom, a single electron travels in an exact circular route around a positively charged nucleus.

According to Bohr's postulates,

$\begin{aligned}\frac{m v^2}{r} & =\frac{1}{4 \pi \epsilon_0} \frac{(Z e)(e)}{r^2} \\m v^2 & =\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r}\end{aligned}$

So, Kinetic Energy

$\begin{aligned}{[K] } & =\frac{1}{2} m v^2 \\K & =\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{2 r}\end{aligned}$

Potential Energy

$\begin{aligned}U & =\frac{1}{4 \pi \epsilon_0} \frac{(Z e)(-e)}{r} \\& =-\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r}\end{aligned}$

Total Energy ,$E=K+U$

Where,

$\begin{aligned}K & =-2 U \\E & =-2 U+U \\& =-2\left[-\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r}\right]+\left[\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r}\right] \\& =\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r}\end{aligned}$

$For n^{\text {th }} orbit, E can be written as E_n So, E_n=\frac{1}{4 \pi \epsilon_0} \frac{Z e^2}{r} For hydrogen atom Z=1 E_n=\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r}$

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