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_B T $ has the same de Broglie wavelength as that of this electron. If this temperature is given by $$ T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B} $$ (where $ h $ is Planck’s constant, $ k_B $ is Boltzmann’s constant, $ m_N $ is the mass of the neutron, and $ a_0 $ is the Bohr radius), then the value of $ \alpha $ is ___.

Hint:

Always express both forms of a physical quantity symbolically, compare with the given, and solve for the unknown. For de Broglie wavelength problems, equating expressions often leads to simplification via squaring.

Answer 1

Correct Answer: 72

Step 1: De Broglie wavelength of the electron The electron is in the \( n = 3 \) orbit of a hydrogen-like atom with atomic number \( Z \). In the Bohr model, the radius of the \( n \)-th orbit is: \[ r_n = \frac{n^2 a_0}{Z} \] For \( n = 3 \): \[ r_3 = \frac{3^2 a_0}{Z} = \frac{9 a_0}{Z} \] where \( a_0 \) is the first Bohr radius of the hydrogen atom (\( Z = 1 \)). The velocity of the electron in the \( n \)-th orbit is: \[ v_n = \frac{Z e^2}{4 \pi \epsilon_0 \hbar n} \] For \( n = 3 \): \[ v_3 = \frac{Z e^2}{12 \pi \epsilon_0 \hbar} \] Substitute \( \hbar = \frac{h}{2\pi} \): \[ v_3 = \frac{Z e^2 \cdot 2\pi}{12 \pi \epsilon_0 h} = \frac{Z e^2}{6 \epsilon_0 h} \] Momentum of the electron: \[ p_e = m_e v_3 = m_e \frac{Z e^2}{6 \epsilon_0 h} \] De Broglie wavelength of the electron: \[ \lambda_e = \frac{h}{p_e} = \frac{6 \epsilon_0 h^2}{m_e Z e^2} \] Alternatively, using the Bohr quantization: \[ n \lambda_e = 2 \pi r_n \Rightarrow 3 \lambda_e = 2 \pi \left( \frac{9 a_0}{Z} \right) \] \[ \lambda_e = \frac{6 \pi a_0}{Z} \] Step 2: De Broglie wavelength of the neutron The neutron has thermal energy \( k_B T \). Using kinetic energy: \[ k_B T = \frac{1}{2} m_N v_N^2 \Rightarrow v_N = \sqrt{\frac{2 k_B T}{m_N}} \] Momentum of the neutron: \[ p_N = \sqrt{2 m_N k_B T} \] De Broglie wavelength: \[ \lambda_N = \frac{h}{p_N} = \frac{h}{\sqrt{2 m_N k_B T}} \] Step 3: Equate the de Broglie wavelengths \[ \lambda_e = \lambda_N \Rightarrow \frac{6 \epsilon_0 h^2}{m_e Z e^2} = \frac{h}{\sqrt{2 m_N k_B T}} \] \[ \frac{6 \epsilon_0 h}{m_e Z e^2} = \frac{1}{\sqrt{2 m_N k_B T}} \Rightarrow \sqrt{2 m_N k_B T} = \frac{m_e Z e^2}{6 \epsilon_0 h} \] Squaring both sides: \[ 2 m_N k_B T = \left( \frac{m_e Z e^2}{6 \epsilon_0 h} \right)^2 = \frac{m_e^2 Z^2 e^4}{36 \epsilon_0^2 h^2} \] \[ k_B T = \frac{m_e^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N} \] Step 4: Use the given temperature expression \[ T = \frac{Z^2 h^2}{a \pi^2 a_0^2 m_N k_B} \Rightarrow k_B T = \frac{Z^2 h^2}{a \pi^2 a_0^2 m_N} \] Equating both expressions: \[ \frac{m_e^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N} = \frac{Z^2 h^2}{a \pi^2 a_0^2 m_N} \] Cancel \( Z^2 \) and \( m_N \): \[ \frac{m_e^2 e^4}{72 \epsilon_0^2 h^2} = \frac{h^2}{a \pi^2 a_0^2} \Rightarrow \frac{m_e^2 e^4}{72 \epsilon_0^2 h^4} = \frac{1}{a \pi^2 a_0^2} \] \[ a \pi^2 a_0^2 = \frac{72 \epsilon_0^2 h^4}{m_e^2 e^4} \] Step 5: Substitute the Bohr radius \( a_0 \) \[ a_0 = \frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2} = \frac{\epsilon_0 h^2}{\pi m_e e^2} \Rightarrow a_0^2 = \frac{\epsilon_0^2 h^4}{\pi^2 m_e^2 e^4} \] Substitute into the earlier expression: \[ a \pi^2 \cdot \frac{\epsilon_0^2 h^4}{\pi^2 m_e^2 e^4} = \frac{72 \epsilon_0^2 h^4}{m_e^2 e^4} \Rightarrow a = 72 \] Step 6: Verify with the alternative electron wavelength Using: \[ \lambda_e = \frac{6 \pi a_0}{Z}, \quad \lambda_N = \frac{h}{\sqrt{2 m_N k_B T}} \] Equating: \[ \frac{6 \pi a_0}{Z} = \frac{h}{\sqrt{2 m_N k_B T}} \Rightarrow \sqrt{2 m_N k_B T} = \frac{Z h}{6 \pi a_0} \] \[ 2 m_N k_B T = \frac{Z^2 h^2}{36 \pi^2 a_0^2} \Rightarrow k_B T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_N} \] Compare with: \[ \frac{Z^2 h^2}{a \pi^2 a_0^2 m_N} \Rightarrow a = 72 \] Final Answer: \[ \boxed{a = 72} \]