Consider an electron in the n = 3 orbit of a hydrogen-like atom with atomic number Z. At absolute te

Q: Why is the kinetic energy of the neutron taken as $k_B T$ instead of $\frac{3}{2} k_B T$?

In many standard physics problems involving thermal neutrons, the 'thermal energy' is often approximated or defined simply as $k_B T$ for the sake of order-of-magnitude calculations or specific problem conventions, as explicitly stated in the problem text 'thermal energy $k_B T$'.

Q: How do I remember the formula for the velocity of an electron in Bohr's model?

You can quickly derive it by equating the electrostatic force to the centripetal force ($\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{Ze^2}{r^2}$) and using the angular momentum quantization condition ($mvr = \frac{nh}{2\pi}$).

Q: Why did we square the Bohr radius $a_0$?

Squaring $a_0$ allows us to create a term $\frac{h^4 \epsilon_0^2}{\pi^2 m_e^2 e^4}$, which perfectly matches and substitutes the messy $\frac{m_e^2 e^4}{\epsilon_0^2}$ part in our temperature equation, simplifying it to the required format.

Q: Does the mass of the electron cancel out in the final expression?

Yes, indirectly! By substituting the Bohr radius $a_0$, which itself contains the mass of the electron, the explicit $m_e$ term is absorbed into $a_0$, leaving the final expression independent of $m_e$ explicitly.

Q: What if the electron was in the $n=2$ orbit?

If $n=2$, the $n^2$ term in the denominator would be $4$. The denominator would then have $8 \times 4 = 32$, making $\alpha = 32$.

Question

Consider an electron in the $n = 3$ orbit of a hydrogen-like atom with atomic number $Z$. At absolute temperature $T$, a neutron having thermal energy $k_BT$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T = \frac{Z^2h^2}{\alpha \pi^2 a_0^2 m_N k_B}$, (where $h$ is the Planck's constant, $k_B$ is the Boltzmann constant, $m_N$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom) then the value of $\alpha$ is ___

Sigma Percentile · Accepted Answer

The problem asks us to find the value of $\alpha$ by equating the de Broglie wavelength of an electron in the $n=3$ orbit of a hydrogen-like atom to that of a thermal neutron. This is a beautiful synthesis of Bohr's atomic model, de Broglie's wave-particle duality, and the kinetic theory of gases.

### Analyzing the Setup

First, let's look at the electron. According to Bohr's model, the velocity of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by:
$$v = \frac{Ze^2}{2\epsilon_0 nh}$$

The de Broglie wavelength $\lambda_e$ associated with this moving electron is the ratio of Planck's constant $h$ to its momentum $m_e v$:
$$\lambda_e = \frac{h}{m_e v}$$

Now, let's consider the neutron. It is at an absolute temperature $T$, which means its thermal kinetic energy is $K = k_B T$. The momentum of the neutron can be expressed in terms of its kinetic energy as $p_N = \sqrt{2m_N K} = \sqrt{2m_N k_B T}$. Therefore, its de Broglie wavelength $\lambda_N$ is:
$$\lambda_N = \frac{h}{\sqrt{2m_N k_B T}}$$

### The Master Equation

The core condition given in the problem is that these two wavelengths are equal:
$$\lambda_e = \lambda_N$$

This implies that their momenta must be equal:
$$m_e v = \sqrt{2m_N k_B T}$$

To find the temperature $T$, we square both sides and isolate $T$:
$$T = \frac{m_e^2 v^2}{2m_N k_B}$$

Now, we substitute the expression for the electron's velocity $v$ into this equation:
$$T = \frac{m_e^2}{2m_N k_B} \left( \frac{Ze^2}{2\epsilon_0 nh} ight)^2$$
$$T = \frac{m_e^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N k_B}$$

### Final Calculation

We are given that the electron is in the $n = 3$ orbit. Substituting $n = 3$ into our equation gives:
$$T = \frac{m_e^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N k_B}$$

The final expression in the problem is given in terms of the first Bohr radius of hydrogen, $a_0$. Let's recall the formula for $a_0$:
$$a_0 = \frac{h^2 \epsilon_0}{\pi m_e e^2}$$

By squaring this formula, we can find a substitution for the mass and charge terms in our temperature equation:
$$a_0^2 = \frac{h^4 \epsilon_0^2}{\pi^2 m_e^2 e^4} \implies \frac{m_e^2 e^4}{\epsilon_0^2} = \frac{h^4}{\pi^2 a_0^2}$$

Substituting this back into our equation for $T$:
$$T = \frac{Z^2}{72 h^2 m_N k_B} \left( \frac{h^4}{\pi^2 a_0^2} ight)$$
$$T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_N k_B}$$

Comparing this derived expression with the given formula $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$, we can clearly see that:
**$\alpha = 72$**

Enter Numerical Value:

Visualized Solution

$v_{n} = \frac{Z e ^{2}}{2 ϵ _{0} nh}$

$λ_{e} = \frac{h}{m _{e} v}$

$λ_{N} = \frac{h}{2 m _{N} k _{B} T}$

$λ_{e} = λ_{N}$

$T = \frac{m _{e}^{2} Z ^{2} e ^{4}}{8 ϵ _{0}^{2} n ^{2} h ^{2} m _{N} k _{B}}$

$n = 3$

$a_{0} = \frac{h ^{2} ϵ _{0}}{π m _{e} e ^{2}}$

$α = 72$

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

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λN​=2mN​kB​T​h​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=8ϵ02​n2h2mN​kB​me2​Z2e4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n=3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a0​=πme​e2h2ϵ0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α=72