In the Bohr model of the hydrogen atoms
(a) the radius of the nth orbit is proportional n^2.
(b) the

Q: Why is the total energy of the electron negative?

The negative sign indicates that the electron is bound to the nucleus. Energy must be supplied to the electron to free it from the atom (to reach zero total energy at an infinite distance).

Q: Does the velocity of the electron increase or decrease in higher orbits?

The velocity decreases. According to Bohr's model, $v_n \propto 1/n$. As the electron moves to higher orbits, it moves slower.

Q: Why is the magnitude of potential energy exactly twice the kinetic energy?

This is a direct consequence of the Virial Theorem for a system bound by an inverse-square force law (like the Coulomb force). For such systems, $2 \times KE + PE = 0$, which means $|PE| = 2 \times KE$.

Q: What happens to the energy difference between consecutive orbits as $n$ increases?

The energy difference decreases. Since $E_n \propto -1/n^2$, the energy levels become closer together as $n$ gets larger, eventually converging at $E = 0$.

Q: Is Bohr's model applicable to multi-electron atoms?

No, Bohr's model only works accurately for hydrogen and hydrogen-like ions (like $\text{He}^+$, $\text{Li}^{2+}$) that have only one electron. It fails to account for electron-electron repulsions in multi-electron atoms.

Question

In the Bohr model of the hydrogen atoms

Options:

(A) the radius of the $n^{	ext{th}}$ orbit is proportional $n^2$.

(B) the total energy of the electron in the $n^{	ext{th}}$ orbit is inversely proportional to $n$.

(C) the angular momentum of the electron in an orbit is an integral multiple of $h/\pi$.

(D) the magnitude of the potential energy of the electron in any orbit is greater than its kinetic energy.

Sigma Percentile · Accepted Answer

The Bohr model of the hydrogen atom is a foundational concept in modern physics, bridging classical mechanics and quantum theory. In this problem, we will evaluate four statements regarding the properties of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom. Let's break down the physics behind each option.

### Analyzing the Radius of the Orbit

According to Bohr's postulates, the electrostatic force between the proton and the electron provides the necessary centripetal force for circular motion. Combined with the quantization of angular momentum, we can derive the radius of the $n^{	ext{th}}$ orbit:
$$r_n = \frac{\epsilon_0 h^2 n^2}{\pi m e^2 Z}$$

Notice the relationship between the radius $r_n$ and the principal quantum number $n$. The radius is directly proportional to the square of $n$:
$$r_n \propto n^2$$

This means that as the electron jumps to higher energy levels, the orbits become significantly larger, scaling quadratically. Therefore, the first statement is **correct**.

### The Total Energy of the Electron

The total energy of the electron is the sum of its kinetic and potential energies. Using the derived radius, the total energy in the $n^{	ext{th}}$ orbit is given by:
$$E_n = -\frac{m e^4 Z^2}{8 \epsilon_0^2 h^2 n^2}$$

Here, the total energy is inversely proportional to the square of the principal quantum number $n$:
$$E_n \propto \frac{1}{n^2}$$

The statement claims that the energy is inversely proportional to $n$, which is a common misconception. Because of the $n^2$ dependence, the second statement is **incorrect**.

### Quantization of Angular Momentum

Bohr's most revolutionary idea was the quantization of angular momentum. He proposed that an electron can only orbit in specific stable circular paths where its angular momentum $L$ is an integral multiple of $\frac{h}{2\pi}$:
$$L = \frac{nh}{2\pi}$$

The option states that the angular momentum is an integral multiple of $\frac{h}{\pi}$. This is missing the crucial factor of $2$ in the denominator. Thus, the third statement is **incorrect**.

### Potential vs. Kinetic Energy

Let's compare the magnitudes of the potential and kinetic energies of the electron. The potential energy $PE$ of the electron-proton system is:
$$PE = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$$

The kinetic energy $KE$ of the orbiting electron is:
$$KE = \frac{1}{8\pi\epsilon_0}\frac{e^2}{r}$$

Taking the absolute value (magnitude) of the potential energy, we get:
$$|PE| = \frac{1}{4\pi\epsilon_0}\frac{e^2}{r}$$

Comparing this with the kinetic energy, it is evident that:
$$|PE| = 2 	imes KE$$

Since the magnitude of the potential energy is exactly twice the kinetic energy, it is always greater than the kinetic energy in any orbit. Therefore, the fourth statement is **correct**.

**Conclusion:** The correct statements are (a) and (d).

Animated Solution for Physics - Atoms and Nuclei: In the Bohr model of the hydrogen atoms

Select Answer:

Visualized Solution

$r_{n} \propto n^{2}$

$E_{n} \propto \frac{1}{n ^{2}}$

$L = \frac{nh}{2 π}$

$∣ P E ∣ > K E$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Analyzing the Radius of the Orbit

The Total Energy of the Electron

Quantization of Angular Momentum

Potential vs

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