Question 31: Dual Nature of Matter and Radiation

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 $k_{\mathrm{B}} T$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_{\mathrm{N}} k_{\mathrm{B}}}$, (where $h$ is the Planck's constant, $k_B$ is the Boltzmann constant, $m_{\mathrm{N}}$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom) then the value of $\alpha$ is ______.

Accepted Answer

Here, we will first find the de Broglie wavelength of the electron in the $n=3$ orbit of a hydrogen-like atom with atomic number $Z$. Then we will find the de Broglie wavelength of the Neutron having thermal energy $k_B T$. Thereafter, we will equate both the wavelengths as they are the same (given)
de Broglie wavelength of the electron:
Radius in the $n$-th orbit,
$$ \begin{aligned} & r_n=\frac{n^2 a_0}{Z} \ & n=3 	ext { So } r=\frac{3^2 a_0}{Z}=\frac{9 a_0}{Z} \end{aligned} $$
Using Bohr quantization :
$$ \begin{aligned} & m_e v r=n \hbar \ \Rightarrow & m_e v \cdot \frac{n^2 a_0}{Z}= \frac{nh}{2 \pi} \ \Rightarrow & p=m_e v=\frac{\hbar h}{2 \pi} \cdot \frac{Z}{n^2 a_0}=\frac{Z h}{2 \pi n a_0} \ & n=3 	ext { so } p=\frac{Z h}{6 \pi a_0} \end{aligned} $$
de Broglie wavelength, $\lambda_e=\frac{h}{p}=\frac{h}{\frac{2 h}{6 \pi q_0}}$
$$ \begin{aligned} &\Rightarrow \lambda_e=\frac{6 \pi a_0}{Z}\ &	ext { Neutron's de Broglie wavelength : }\ &E=k_B T 	ext { (given) } \end{aligned} $$
$$ \begin{aligned} &\begin{aligned} p & =\sqrt{2 m_N E}=\sqrt{2 m_N k_B T} \ \lambda_n & =\frac{h}{p}=\frac{h}{\sqrt{2 m_N k_B T}} \end{aligned}\ &	ext { Now, equating the wavelengths, }\ &\begin{aligned} & \frac{6 \pi a_0}{Z}=\frac{h}{\sqrt{2 m_N k_B T}} \ \Rightarrow & \sqrt{2 m_N k_B T}=\frac{Z h}{6 \pi a_0} \ \Rightarrow & 2 m_N k_B T=\frac{Z^2 h^2}{36 \pi^2 a_0^2} \ \Rightarrow & T=\frac{Z^2 h^2}{72 m_N k_B \pi^2 a_0^2} \ & T=\frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B} 	ext { (given) } \end{aligned} \end{aligned} $$
$$ 	ext { So, } \alpha=72 $$

Dual Nature of Matter and Radiation

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