Imagine an atom made up of proton and a hypothetical particle of double the mass of the electron but

Q: Why does the Rydberg constant depend on the mass of the electron?

In the derivation of the Bohr model, the energy levels depend on the mass of the orbiting particle because the centripetal force required to keep the particle in orbit is directly proportional to its mass. This mass term gets bundled into what we call the Rydberg constant.

Q: Why is the first excited state $n=2$ and not $n=1$?

The lowest possible energy state (the ground state) is $n=1$. Any state above the ground state is considered 'excited'. Therefore, the first step up from the ground state is $n=2$, making it the first excited state.

Q: How do we know that the longest wavelength corresponds to the transition from $n=3$ to $n=2$?

The energy of a photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$). To get the maximum (longest) wavelength, you need the minimum energy. The smallest energy gap ending at $n=2$ is from the level immediately above it, which is $n=3$.

Q: What would happen if the mass of the particle was halved instead of doubled?

If the mass was halved, the Rydberg constant would also be halved. The energy levels would become $E'_n = -\frac{Rhc}{2n^2}$, and the energy differences between levels would be exactly half of those in a standard hydrogen atom.

Q: Does the charge of the hypothetical particle affect the energy levels?

Yes, the energy levels depend on the square of the charge of the orbiting particle and the nucleus. However, the problem states that the hypothetical particle has the same charge as an electron, so the charge factor remains unchanged in our calculations.

Question

Imagine an atom made up of proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength $\lambda$ (given in terms of the Rydberg constant $R$ for the hydrogen atom) equal to

Options:

(A) $\frac{9}{5R}$

(B) $\frac{36}{5R}$

(C) $\frac{18}{5R}$

(D) $\frac{4}{R}$

Sigma Percentile · Accepted Answer

The Bohr model of the hydrogen atom is a beautiful framework that elegantly explains the discrete energy levels of an electron orbiting a proton. But what happens if we tweak the rules of the universe just a little bit?

In this fascinating problem, we are asked to imagine a hypothetical atom. It still has a proton at its center, and it still has a negatively charged particle orbiting it. However, this new particle is twice as massive as a standard electron! Let's dive into how this single change ripples through the physics of the atom.

### The Mass Dependency of Energy Levels

To understand the impact of a heavier orbiting particle, we first need to look at the standard formula for the energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen atom:
$$E_n = -\frac{Rhc}{n^2}$$
Here, $R$ is the Rydberg constant, $h$ is Planck's constant, and $c$ is the speed of light.

Now, here is the crucial insight: the Rydberg constant $R$ is not just a random number; it is directly proportional to the mass of the revolving particle ($R \propto m$). Because the mass of our hypothetical particle is $2m_e$ (double the mass of an electron), the new Rydberg constant effectively doubles. Consequently, the energy of every single orbit is doubled!
$$E'_n = 2E_n = -\frac{2Rhc}{n^2}$$

### Finding the Longest Wavelength

The problem asks us to find the **longest wavelength** photon emitted when the particle transitions to the **first excited level**.

Let's break this down:
1. **First Excited Level:** The ground state is $n = 1$, which means the first excited level is $n = 2$.
2. **Longest Wavelength:** According to the Planck-Einstein relation ($E = \frac{hc}{\lambda}$), wavelength is inversely proportional to energy. Therefore, the longest wavelength corresponds to the **minimum energy** transition.
3. **Minimum Energy Transition:** To emit the least amount of energy while landing on $n = 2$, the particle must fall from the energy level immediately above it, which is $n = 3$.

### The Final Calculation

Now that we know the transition is from $n = 3$ to $n = 2$, we can calculate the energy difference $\Delta E$:
$$\Delta E = E'_3 - E'_2$$
Substitute our modified energy formula into the equation:
$$\Delta E = \left(-\frac{2Rhc}{3^2}ight) - \left(-\frac{2Rhc}{2^2}ight)$$
$$\Delta E = 2Rhc \left(\frac{1}{4} - \frac{1}{9}ight)$$

Finding a common denominator (which is 36):
$$\Delta E = 2Rhc \left(\frac{9 - 4}{36}ight) = 2Rhc \left(\frac{5}{36}ight) = \frac{5Rhc}{18}$$

Finally, we equate this energy difference to the energy of the emitted photon to find its wavelength $\lambda_{	ext{max}}$:
$$\frac{hc}{\lambda_{	ext{max}}} = \frac{5Rhc}{18}$$

The $hc$ terms cancel out perfectly. Taking the reciprocal of both sides, we arrive at our final answer:
$$\lambda_{	ext{max}} = \frac{18}{5R}$$

By simply doubling the mass of the orbiting particle, we scaled the entire energy landscape of the atom, leading us to this elegant result!

Select Answer:

Visualized Solution

$E_{n} = - \frac{R h c}{n ^{2}}$

$R \propto m$

$E_{n}^{'} = - \frac{2 R h c}{n ^{2}}$

$First Excited Level ⟹ n = 2$

$λ_{max} ⟹ Δ E_{min}$

$Δ E = E_{3}^{'} - E_{2}^{'}$

$Δ E = 2 R h c (\frac{1}{4} - \frac{1}{9})$

$λ_{max} = \frac{18}{5 R}$

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

The Mass Dependency of Energy Levels

Finding the Longest Wavelength

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En​=−n2Rhc​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } En′​=−n22Rhc​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Excited Level⟹n=2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λmax​⟹ΔEmin​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=E3′​−E2′​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔE=2Rhc(41​−91​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λmax​=5R18​