Question

What will be the uncertainty in velocity of an electron
when the uncertainty in its position is 1000 A?
3 5.79 x 102 m s-1 2) 5.79 108 m s-1
3) 5.79 x 104 m s- 4) 5.79 × 10-10 m s-1​

Answer 1

Solution

Uncertainty in velocity of an electron is calculated by using Heisenberg's Uncertainty Principle

From the Question,

$\sf{\Delta{x} = {10}^{-7}}$

We know that,

$\sf{Mass \ of \ an \ electron,m = 9.1 × {10}^{-31}}$

Heisenberg's Uncertainty Principle is given by,

$\sf{\Delta{x} \times \Delta{p} \geqslant \frac{h}{4 \pi}}$

$\implies \sf{\Delta{x} \times m.\Delta{v} \geqslant \frac{h}{4 \pi}}$

$\implies \: \sf{10 {}^{ - 7} \times 9.1 \times {10}^{ - 31} \Delta{v} \geqslant \frac{6.25 \times 10 {}^{ - 34} }{4 \times 3.14}} \\ \implies \: \sf{\Delta{v} = \frac{6.25 \times {10}^{ - 34} }{4 \times 3.14 \times 9.1 \times {10}^{ - 38} } } \\ \\ \implies \: \boxed{\sf{\Delta{v} = 5.79 \times {10}^{2}ms {}^{ - 1}}}$

The uncertainty in velocity of the electron is 5.79×10² m/s

Answer 2

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

${ \triangle x = 1000 Angstrom}$

${ m =9.1 \times 10^-31 kg }$

${ \triangle v = (to find)}$

$\huge{\underline{Explanation:-}}$

From Heisenberg's uncertainty principle.

$\huge{\boxed{\boxed{ \triangle x \times \triangle v = \frac{h}{4\pi \: m}}}}$

Substituting the values

${ \triangle v = \frac{6.6 \times {10}^{ - 34} }{4 \times 3.14 \times 9.1 \times {10}^{ - 31} \times 1000 \times {10}^{ - 10}}}$

${ \triangle v = \frac{6.6 \times {10}^{ - 34} }{4 \times 3.14 \times 9.1 \times {10}^{ - 31 - 10 + 3}}}$

solving it

${ \triangle v = \frac{6.6 \times {10}^{ 4} }{4 \times 3.14 \times 9.1}}$

${ \triangle v = 0.05789 \times 10^4 m/s}$

${ \triangle v = 5.789 \times 10^2 m/s}$

$\huge{\boxed{\boxed{ \triangle v = 5.789 \times 10^2 m/s}}}$

So, uncertainty in its velocity is 5.789 ×10² (option --- 1).