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Atomic Structure

Section: Chemistry

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Question : 3 of 32

Marks: +1, -0

The uncertainty in the velocities of two particles A and B are 0.03 and 0.01

\mathrm{m}\,\mathrm{s}^{-1}

respectively. The mass of

B

is four times to the mass of

A

. The ratio of uncertainties in their positions is

[AP EAPCET 23 May 2025 S2]

$\frac{3}{4}$
$\frac{16}{9}$
$\frac{9}{16}$
$\;{9}/{16}$

Solution:

Uncertainty principle for particle

\Delta x_A (m_A \cdot \Delta v_A) = \frac{h}{4\pi} \; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)

;

B

For particle

\Delta x_B (m_B \cdot \Delta v_B) = \frac{h}{4\pi} \; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)

,

\frac{\Delta x_A}{\Delta x_B} = \frac{m_B \cdot \Delta v_B}{m_A \cdot \Delta v_A}

Divide Eq. (i) by Eq. (ii)

m_B = 4 m_A, \Delta v_A = 0.03

Now substitute the given value

\Delta v_B = 0.01

\frac{\Delta x_A}{\Delta x_B} = \frac{0.04}{0.03} = \frac{4}{3}

\frac{4}{3}

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