Ncert Exercises & Solutions

Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using the Heisenberg uncertainty principle $(∆ x \times ∆ p \approx h) .$ You can assume the uncertainty in position $∆$ x as 1 nm. Assuming $p \approx ∆ p,$ find the energy of the electron in electronvolts.

Hint: Use the Heisenberg uncertainty principle.

Step 1: Find the momentum and energy of the electron.

\begin{aligned} ∴ Δp & = \frac{h}{2 πΔx} \\ \Rightarrow = \frac{6 .62 \times 10^{- 34} JS}{2 \times (22 / 7) \times (10^{- 9}) m} \\ = 1 .05 \times 10^{- 25} kgm / s \\ Energy, E & = \frac{p^{2}}{2 m} = \frac{(Δp)^{2}}{2 m} [∵ p \approx ∆ p] \\ = \frac{{(1 .05 \times 10^{- 25})}^{2}}{2 \times 9 .1 \times 10^{- 31}} J \\ = \frac{{(1 .05 \times 10^{- 25})}^{2}}{2 \times 9 .1 \times 10^{- 31} \times 1 .6 \times 10^{- 19}} eV \\ = 3 .8 \times 10^{- 2} eV \end{aligned}

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