A particle of charge equal to that of an electron -e, and mass 208 times of the mass of the electron

Q: Why do we assume the nucleus is at rest?

The problem states to take the mass of the nucleus as infinite. In reality, the nucleus is much heavier than the mu-meson, so the center of mass of the system is essentially at the center of the nucleus, making it a valid approximation to consider it at rest.

Q: How does the mass of the orbiting particle affect the Bohr radius?

The Bohr radius is inversely proportional to the mass of the orbiting particle. Since the mu-meson is 208 times heavier than an electron, its orbit will be significantly closer to the nucleus compared to an electron in the same energy level.

Q: Why does the energy scale with the mass of the particle?

The total energy of the orbiting particle is proportional to its mass. A heavier particle like the mu-meson will have a much deeper (more negative) energy well, meaning it is bound much more tightly to the nucleus than an electron would be.

Q: Can we use the standard Rydberg formula directly here?

You can, but you must modify the Rydberg constant. The standard Rydberg constant $R$ is for an electron. For a mu-meson, the effective Rydberg constant becomes $R_\mu = R \times (m_\mu / m_e) \times Z^2$. It's often easier to scale the energy directly from the $-13.6 \text{ eV}$ value.

Q: Why did we use $12375$ to find the wavelength?

The value $12375$ (or sometimes $12400$) is a handy conversion factor for $hc$ when energy is in electron-volts ($\text{eV}$) and wavelength is in Angstroms ($\text{\AA}$). It saves time compared to plugging in the standard SI values for $h$, $c$, and $e$.

Question

A particle of charge equal to that of an electron $-e$, and mass $208$ times of the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge $+3e$. (Take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system,
(a) derive an expression for the radius of the $n^{	ext{th}}$ Bohr orbit.
(b) find the value of $n$ for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.
(c) find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit to the first orbit. (Rydberg's constant = $1.097 	imes 10^7 	ext{ m}^{-1}$)

Sigma Percentile · Accepted Answer

### Unveiling the Mu-Meson Atom: A Journey Through Bohr's Model

Imagine a tiny, exotic solar system. Instead of a standard electron orbiting a single proton, we have a massive mu-meson—a particle $208$ times heavier than an electron—orbiting a nucleus with a charge of $+3e$. This is a fascinating thought experiment that tests our understanding of Bohr's atomic model. Let's break down the physics step-by-step.

### The Master Equation: Balancing Forces

For any particle to move in a stable circular orbit, it requires a centripetal force. In our atomic system, this force is provided entirely by the electrostatic attraction between the positively charged nucleus and the negatively charged mu-meson.

We can write this balance as:

$$\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{(Ze)(e)}{r^2}$$

Here, $Z=3$ for our nucleus, and $m = 208 m_e$. But we have two unknowns: the velocity $v$ and the radius $r$. To solve this, we need Bohr's famous quantization rule, which states that the angular momentum must be an integral multiple of $h/2\pi$:

$$mvr = \frac{nh}{2\pi}$$

By substituting $v$ from the angular momentum equation into our force balance equation, we can derive the general expression for the radius of the $n^{	ext{th}}$ orbit:

$$r = \frac{n^2 h^2 \varepsilon_0}{Z \pi m e^2}$$

### Finding the Mu-Meson's Radius

Now, let's tailor this equation to our specific mu-meson system. We substitute $Z = 3$ and $m = 208 m_e$ into the radius formula:

$$r_n = \frac{n^2 h^2 \varepsilon_0}{3 \pi (208 m_e) e^2} = \frac{n^2 h^2 \varepsilon_0}{624 \pi m_e e^2}$$

This is our answer for part (a). Notice how the large mass and higher nuclear charge drastically shrink the orbit compared to a standard hydrogen atom!

### Equating with the Hydrogen Atom

In part (b), we are asked to find the quantum number $n$ where this mu-meson's orbit matches the size of the first Bohr orbit of a hydrogen atom. The radius of hydrogen's first orbit ($Z=1, m=m_e, n=1$) is:

$$r_H = \frac{h^2 \varepsilon_0}{\pi m_e e^2}$$

Equating our mu-meson radius $r_n$ to $r_H$:

$$\frac{n^2 h^2 \varepsilon_0}{624 \pi m_e e^2} = \frac{h^2 \varepsilon_0}{\pi m_e e^2}$$

All the constants cancel out beautifully, leaving us with a simple relation:

$$n^2 = 624$$

Taking the square root, we find that $n \approx 25$. This means the mu-meson has to be in its $25^{	ext{th}}$ excited state just to have an orbit as large as a hydrogen atom's ground state!

### The Energy Leap and Emitted Radiation

Finally, let's tackle part (c). We need to find the wavelength of the photon emitted when the mu-meson jumps from $n=3$ to $n=1$. First, we need the energy expression. The total energy of an orbiting particle is:

$$E_n = -\frac{Z^2 m e^4}{8 \varepsilon_0^2 h^2 n^2}$$

Instead of calculating this from scratch, we can relate it to the known ground state energy of hydrogen, $E_0 = -13.6 	ext{ eV}$. The energy scales with $Z^2$ and the mass $m$:

$$E_n = \frac{Z^2 (m/m_e)}{n^2} E_0 = \frac{3^2 	imes 208}{n^2} (-13.6 	ext{ eV}) = -\frac{25459.2}{n^2} 	ext{ eV}$$

The energy difference $\Delta E$ for the transition from $n=3$ to $n=1$ is:

$$\Delta E = E_3 - E_1 = -25459.2 \left( \frac{1}{3^2} - \frac{1}{1^2} ight) = 25459.2 	imes \frac{8}{9} = 22630.4 	ext{ eV}$$

To find the wavelength $\lambda$ in Angstroms, we use the handy conversion $hc \approx 12375 	ext{ eV \AA}$:

$$\lambda = \frac{hc}{\Delta E} = \frac{12375}{22630.4} \approx 0.546 	ext{ \AA}$$

This incredibly short wavelength falls deep into the X-ray spectrum, a direct consequence of the mu-meson's large mass and the strong nuclear charge binding it tightly to the nucleus.

Visualized Solution

$System Setup$

$Force Balance$

$Bohr’s Quantization$

$Radius of n^{th} Orbit$

$Radius for Mu-Meson$

$Equating with H-atom$

$Solving for n$

$Energy of n^{th} Orbit$

$Energy in terms of Rydberg$

$Energy Difference$

$Wavelength of Radiation$

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

Unveiling the Mu-Meson Atom

The Master Equation

Finding the Mu-Meson's Radius

Equating with the Hydrogen Atom

The Energy Leap and Emitted Radiation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bohr’s Quantization

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius of nth Orbit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius for Mu-Meson

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating with H-atom

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for n

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of nth Orbit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy in terms of Rydberg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Difference

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Wavelength of Radiation