Q2 / 160 · AP EAMCET 2023 (18 May Shift 1) · Chemistry

If the uncertainty in velocity of electron $(\Delta \mathrm{v})$ is $0.1 \mathrm{~m} / \mathrm{s}$, the uncertainty in its position $(\Delta x)$ is (given: $m_e=9.1 \times 10^{-31} \mathrm{~kg}$ )

  • A
    $2.02 \times 10^{-4} \mathrm{~m}$
  • B
    $4.04 \times 10^{-4} \mathrm{~m}$
  • Correct answer
    C
    $5.79 \times 10^{-4} \mathrm{~m}$
  • D
    $8.42 \times 10^{-4} \mathrm{~m}$

Solution

$\Delta x \cdot \Delta \mathrm{P}=\Delta \mathrm{x} \cdot \mathrm{m} \Delta \mathrm{v}=\frac{\mathrm{h}}{4 \pi}=\frac{6.626 \times 10^{-34}}{4 \times 3.14}$ $\begin{aligned} & =5.27 \times 10^{-35} \\ & \Rightarrow \Delta x=\frac{5.27 \times 10^{-35}}{\left(9.1 \times 10^{-31}\right)(0.1)}=5.79 \times 10^{-4} \mathrm{~m}\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions