Hydrogen-like Atom Challenge

Quantum physics meets thermal de Broglie wavelength

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 \( k_B T \) has the same de Broglie wavelength as that of this electron. If this temperature is given by:

\[ T = \frac{z^2 h^2}{\alpha n^2 a_0^2 m_N k_B} \]

(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 ______

Given: \( n = 3 \)

Answer: \( \alpha = 72 \)

Detailed Solution

Step 1: Find the electron's de Broglie wavelength in the n=3 orbit

For a hydrogen-like atom, the electron's velocity in the nth orbit is:

\[ v_n = \frac{Z e^2}{2 \epsilon_0 h n} \]

The de Broglie wavelength is:

\[ \lambda_e = \frac{h}{m_e v_n} = \frac{2 \epsilon_0 h^2 n}{m_e Z e^2} \]

Step 2: Express the neutron's de Broglie wavelength

The thermal energy of the neutron is \( E = k_B T \), and its momentum is:

\[ p = \sqrt{2 m_N E} = \sqrt{2 m_N k_B T} \]

The de Broglie wavelength is:

\[ \lambda_N = \frac{h}{p} = \frac{h}{\sqrt{2 m_N k_B T}} \]

Step 3: Set the wavelengths equal and solve for T

Setting \( \lambda_e = \lambda_N \):

\[ \frac{2 \epsilon_0 h^2 n}{m_e Z e^2} = \frac{h}{\sqrt{2 m_N k_B T}} \]

Squaring both sides and solving for T:

\[ T = \frac{m_e^2 Z^2 e^4}{8 \epsilon_0^2 h^2 n^2 m_N k_B} \]

Step 4: Relate to Bohr radius and simplify

The Bohr radius is \( a_0 = \frac{\epsilon_0 h^2}{\pi m_e e^2} \), so:

\[ T = \frac{Z^2 h^2}{72 n^2 a_0^2 m_N k_B} \]

Comparing with the given form \( T = \frac{Z^2 h^2}{\alpha n^2 a_0^2 m_N k_B} \), we get:

\[ \alpha = 72 \]

de Broglie Wavelength Explorer

Hydrogen-like Atom (Z = 1)

Electron in n = 3 orbit

Neutron at T = 0 K

Atomic Number (Z):

1

Principal Quantum Number (n):

3

Electron wavelength (orbit): 0 m

Neutron wavelength (thermal): 0 m

Required temperature: 0 K

💡 Did you know? The de Broglie wavelength concept bridges classical and quantum physics. For macroscopic objects, the wavelength is so small it's negligible, but for electrons and neutrons at appropriate energies, wave behavior becomes significant. This is why electron microscopes can achieve much higher resolution than optical microscopes!