Question Details

22. 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 the temperature is given by $T = \frac{Z^{2} h^{2}}{α π a_{0}^{2} m_{N} k B}$ , (where $h$ is Planck’s constant, $k B$ is Boltzmann constant, $m_{n}$ is the mass of neutron and $a_{0}$ is the first Bohr radius of hydrogen atom), then the value of $α$ is ______.

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Correct Answer :

Solution :

The correct answer is 72.

To find the value of $α$ , we equate the de Broglie wavelength of the electron in the given orbit to the de Broglie wavelength of the neutron.

Step 1: Find the de Broglie wavelength of the electron in the Bohr orbit
According to Bohr's quantization of angular momentum for an electron in the $n$ -th orbit:
$m_{e} v r_{n} = \frac{n h}{2 π}$
The momentum of the electron is:
$p_{e} = m_{e} v = \frac{n h}{2 π r_{n}}$
The de Broglie wavelength of the electron is given by:
$λ_{e} = \frac{h}{p_{e}} = \frac{2 π r_{n}}{n}$

For a hydrogen-like atom with atomic number $Z$ , the radius of the $n$ -th orbit is:
$r_{n} = \frac{n^{2} a_{0}}{Z}$
Substituting $r_{n}$ into the equation for $λ_{e}$ :
$λ_{e} = \frac{2 π}{n} \cdot \frac{n^{2} a_{0}}{Z} = \frac{2 π n a_{0}}{Z}$

For an electron in the $n = 3$ orbit:
$λ_{e} = \frac{6 π a_{0}}{Z}$

Step 2: Find the de Broglie wavelength of the neutron
The thermal energy of the neutron at temperature $T$ is:
$E = k_{B} T$
The momentum of the neutron is:
$p_{N} = \sqrt{2 m_{N} k_{B} T}$
The de Broglie wavelength of the neutron is:
$λ_{N} = \frac{h}{p_{N}} = \frac{h}{\sqrt{2 m_{N} k_{B} T}}$

Step 3: Equating both wavelengths to solve for T
Since $λ_{e} = λ_{N}$ :
$\frac{6 π a_{0}}{Z} = \frac{h}{\sqrt{2 m_{N} k_{B} T}}$
Squaring both sides:
$\frac{36 π^{2} a_{0}^{2}}{Z^{2}} = \frac{h^{2}}{2 m_{N} k_{B} T}$
Rearranging to solve for the absolute temperature $T$ :
$T = \frac{Z^{2} h^{2}}{72 π^{2} a_{0}^{2} m_{N} k_{B}}$