Consider an electron in the orbit of a hydrogen-like atom with atomic number . At absolute temperature , a neutron having thermal energy has the same de Broglie wavelength as that of this electron. If this temperature is given by (where is the Planck'

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 $T = \frac{z^{2} h^{2}}{α π^{2} a_{0}^{2} m_{N} k_{B}},$ (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 $α$ is ______. [2025]

Smart Achievers · Accepted Answer

(72)

From Bohr's postulates,

$m v r = \frac{n h}{2 π}$ $\dots (i)$

and $\frac{m v^{2}}{r} = \frac{K Z e^{2}}{r^{2}}$

$\Rightarrow m v^{2} r = \frac{1}{4 π ε_{0}} Z e^{2}$ $\dots (i i)$

Dividing Eq. (ii) by Eq. (i),

$v = \frac{\frac{Z e^{2}}{4 π ε_{0}}}{\frac{n h}{2 π}} = \frac{Z e^{2}}{2 ε_{0} n h}$

$λ = \frac{h}{m v} = \frac{h}{\sqrt{2 m_{N} K_{B} T}}$

Now putting the above value of $v$ and solving, we get

$T = \frac{m^{2} Z^{2} e^{4}}{8 ε_{0}^{2} n^{2} h^{2} m_{N} K_{B}}$ or, $T = \frac{m^{2} Z^{2} e^{4}}{72 ε_{0}^{2} h^{2} m_{N} K_{B}}$ $[∵ n = 3 given]$

Again dividing Eq. (ii) by Eq. (i), we get

$\frac{1}{m r} = \frac{\frac{Z e^{2}}{4 π ε_{0}}}{\frac{n^{2} h^{2}}{4 π^{2}}}$

$r = \frac{n^{2} h^{2} ε_{0}}{π Z e^{2} m}$ $\Rightarrow a_{0} = \frac{h^{2} ε_{0}}{π e^{2} m}$ $[∵ r = a_{0} n^{2}]$

or, $a_{0}^{2} = \frac{h^{4} ε_{0}^{2}}{π^{2} e^{4} m^{2}}$

$T a_{0}^{2} = \frac{m^{2} Z^{2} e^{4}}{72 ε_{0}^{2} h^{2} m_{N} K_{B}} \cdot \frac{h^{4} ε_{0}^{2}}{π^{2} e^{4} m^{2}}$

or, $T = \frac{h^{2} Z^{2}}{72 π^{2} a_{0}^{2} m_{N} K_{B}}$

Comparing with the given $T = \frac{Z^{2} h^{2}}{α π^{2} a_{0}^{2} m_{N} K_{B}},$

we get $α = 72$

Solution

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