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Assuming an electron is confined to $1 \mathrm{~nm}$ wide region, find the uncertainty in momentum using Heisenberg uncertainty principle $(\Delta \mathrm{x} \times \Delta \mathrm{p} \approx \mathrm{h})$. You can assume the uncertainty in position $\Delta \mathrm{x}$ as $1 \mathrm{~nm}$. Assuming $\mathrm{p} \approx \Delta \mathrm{p}$, find the energy of the electron in electron volts.
Solution:
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Verified Answer
Given, electron revalves in circular path so, $\Delta x$
$$
\begin{aligned}
&\Delta \mathrm{x}=1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \Delta \mathrm{p}=\text { ? } \\
&\text { As } \Delta x^{\prime} \Delta \mathrm{p} \approx \mathrm{h} \\
&\therefore \quad \Delta \mathrm{p}=\frac{\mathrm{h}}{\Delta \mathrm{x}^{\prime}}=\frac{\mathrm{h}}{2 \pi \Delta \mathrm{x}} \quad\left(\because \Delta \mathrm{x}^{\prime}=2 \pi \Delta \mathrm{x}\right) \\
&\left(\because \mathrm{h}=6.62 \times 10^{-34}\right) \\
&=\frac{6.62 \times 10^{-34} \mathrm{JS}}{2 \times(22 / 7)\left(10^{-9}\right)_{\mathrm{m}}}=1.05 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
So energy, $E=\frac{\mathrm{p}^2}{2 \mathrm{~m}}=\frac{(\Delta \mathrm{p})^2}{2 \mathrm{~m}} \quad(\because \mathrm{p} \approx \Delta \mathrm{p})$
$$
=\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}} \mathrm{~J} \quad\left(\because \mathrm{n}=9.1 \times 10^{-31}\right)
$$
$$
\begin{aligned}
\Rightarrow &=\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
&=3.8 \times 10^{-2} \mathrm{eV}
\end{aligned}
$$
$$
\begin{aligned}
&\Delta \mathrm{x}=1 \mathrm{~nm}=10^{-9} \mathrm{~m}, \Delta \mathrm{p}=\text { ? } \\
&\text { As } \Delta x^{\prime} \Delta \mathrm{p} \approx \mathrm{h} \\
&\therefore \quad \Delta \mathrm{p}=\frac{\mathrm{h}}{\Delta \mathrm{x}^{\prime}}=\frac{\mathrm{h}}{2 \pi \Delta \mathrm{x}} \quad\left(\because \Delta \mathrm{x}^{\prime}=2 \pi \Delta \mathrm{x}\right) \\
&\left(\because \mathrm{h}=6.62 \times 10^{-34}\right) \\
&=\frac{6.62 \times 10^{-34} \mathrm{JS}}{2 \times(22 / 7)\left(10^{-9}\right)_{\mathrm{m}}}=1.05 \times 10^{-25} \mathrm{~kg} \mathrm{~m} / \mathrm{s} \\
&
\end{aligned}
$$
So energy, $E=\frac{\mathrm{p}^2}{2 \mathrm{~m}}=\frac{(\Delta \mathrm{p})^2}{2 \mathrm{~m}} \quad(\because \mathrm{p} \approx \Delta \mathrm{p})$
$$
=\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31}} \mathrm{~J} \quad\left(\because \mathrm{n}=9.1 \times 10^{-31}\right)
$$
$$
\begin{aligned}
\Rightarrow &=\frac{\left(1.05 \times 10^{-25}\right)^2}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}} \mathrm{eV} \\
&=3.8 \times 10^{-2} \mathrm{eV}
\end{aligned}
$$
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