JEE MAIN - Chemistry (2022 - 29th July Morning Shift - No. 17)
The minimum uncertainty in the speed of an electron in an one dimensional region of length $$2 \mathrm{a}_{\mathrm{o}}$$ (Where $$\mathrm{a}_{\mathrm{o}}=$$ Bohr radius $$52.9 \,\mathrm{pm}$$) is _________ $$\mathrm{km} \,\mathrm{s}^{-1}$$.
(Given : Mass of electron = 9.1 $$\times$$ 10$$-$$31 kg, Planck's constant h = 6.63 $$\times$$ 10$$-$$34 Js)
Answer
548
Explanation
Heisenberg's uncertainty principle
$$ \begin{aligned} & \Delta \mathrm{x} \times \Delta \mathrm{P}_{\mathrm{x}} \geq \frac{h}{4 \pi} \\ & \Rightarrow 2 \mathrm{a}_{0} \times \mathrm{m} \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \text { (minimum) } \\ & \Rightarrow \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \times \frac{1}{2 a_{0}} \times \frac{1}{m} \\ & =\quad 6.63 \times 10^{-34} \end{aligned} $$
$$ \begin{aligned} &4 \times 3.14 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31} \\ & =548273 \,\mathrm{~ms}^{-1} \\ & =548.273 \,\mathrm{kms}^{-1} \\ & =548 \,\mathrm{kms}^{-1} \end{aligned} $$
$$ \begin{aligned} & \Delta \mathrm{x} \times \Delta \mathrm{P}_{\mathrm{x}} \geq \frac{h}{4 \pi} \\ & \Rightarrow 2 \mathrm{a}_{0} \times \mathrm{m} \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \text { (minimum) } \\ & \Rightarrow \Delta \mathrm{v}_{\mathrm{x}}=\frac{h}{4 \pi} \times \frac{1}{2 a_{0}} \times \frac{1}{m} \\ & =\quad 6.63 \times 10^{-34} \end{aligned} $$
$$ \begin{aligned} &4 \times 3.14 \times 2 \times 52.9 \times 10^{-12} \times 9.1 \times 10^{-31} \\ & =548273 \,\mathrm{~ms}^{-1} \\ & =548.273 \,\mathrm{kms}^{-1} \\ & =548 \,\mathrm{kms}^{-1} \end{aligned} $$
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