Question

calculate uncertainty in position of electron moving with the velocity 300 m/s with 0.001% accuracy.
(mass of electron=9.1×10^-31 kg)
(h=6.625×10^-34 j/s)

Answer 1

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

• Velocity of electron is 300 m/s

• Percentage Uncertainity in it's velocity is 0.001%

• h = 6.625 × 10⁻³⁴J/sec

• Mass of electron = 9.1 × 10⁻³¹ kg

$\huge {\bold {\underline {\underline{To \: find : - }}}}$

• Uncertainity in it's position

$\huge {\bold {\underline {\underline{Solution : - }}}}$

%ΔV is given by ,

$\large {\sf {\underline {\bold {\boxed {\bigstar{ \: | \%\triangle v = \dfrac{ \triangle v}{v} \times 100}}}}}}$

We have ,

• v = 300 m/s
• % Δv = 0.001%

$\sf : \implies \: 0.001 = \dfrac{ \triangle v}{300} \times 100 \\ \\ \sf : \implies \: 0.001 = \dfrac{ \triangle v}{3 \cancel{00}} \times 1 \cancel{00} \\ \\ \sf : \implies \: 0.001 = \dfrac{ \triangle v}{3} \\ \\ \sf: \implies3 \times 0.001 = \triangle v \\ \\ \sf : \implies {\bold {\boxed {\pink {\triangle v = 3 \times {10}^{ - 3} \: m {s}^{ - 1} }}}}$

From Heisenberg's uncertainity principle ,

$\large {\sf {\underline{\bold {\boxed{\bigstar{ \: (\triangle x)(m \triangle v) = \dfrac{h}{4\pi} }}}}}}$

Where ,

• Δx is uncertainity in position
• m is mass
• Δv is uncertainity in velocity
• h is planck's constant

$\sf: \implies \: ( \triangle x)[ (9.1 \times {10}^{ - 31} ( 3 \times {10}^{ - 3})]= \dfrac{6.625 \times {10}^{ - 34} }{4 \times 3.14} \\ \\ \sf: \implies (\triangle x)(27.3 \times {10}^{ - 34} ) = \dfrac{6.625 \times {10}^{ - 34} }{4 \times 3.14} \\ \\ \sf : \implies( \triangle x) = \dfrac{6.625 \times {10}^{ - 34} }{4 \times 3.14 \times 27.3 \times {10}^{ - 34} } \\ \\\sf : \implies (\triangle x) = \dfrac{6.625 \times \cancel{ {10}^{ - 34} }}{342.888 \times \cancel{ {10}^{ - 34} }} \\ \\ \sf : \implies ( \triangle x) = \dfrac{6.625}{342.888} \\ \\ \sf : \implies {\bold {\boxed {\pink{(\triangle x) = 0.019 \: m}}}}$

∴ The uncertainity in position of the electron is 0.019 m

Answer 2

Answer:

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

Given:−

Velocity of electron is 300 m/s

Percentage Uncertainity in it's velocity is 0.001%

h = 6.625 × 10⁻³⁴J/sec

Mass of electron = 9.1 × 10⁻³¹ kg

\huge {\bold {\underline {\underline{To \: find : - }}}}

Tofind:−

Uncertainity in it's position

\huge {\bold {\underline {\underline{Solution : - }}}}

Solution:−

%ΔV is given by ,

\large {\sf {\underline {\bold {\boxed {\bigstar{ \: | \%\triangle v = \dfrac{ \triangle v}{v} \times 100}}}}}}

★∣%△v=

v

△v

×100

We have ,

v = 300 m/s

% Δv = 0.001%

\begin{gathered} \sf : \implies \: 0.001 = \dfrac{ \triangle v}{300} \times 100 \\ \\ \sf : \implies \: 0.001 = \dfrac{ \triangle v}{3 \cancel{00}} \times 1 \cancel{00} \\ \\ \sf : \implies \: 0.001 = \dfrac{ \triangle v}{3} \\ \\ \sf: \implies3 \times 0.001 = \triangle v \\ \\ \sf : \implies {\bold {\boxed {\pink {\triangle v = 3 \times {10}^{ - 3} \: m {s}^{ - 1} }}}}\end{gathered}

:⟹0.001=

300

△v

×100

:⟹0.001=

3

00

△v

×1

00

:⟹0.001=

3

△v

:⟹3×0.001=△v

:⟹

△v=3×10

−3

ms

−1

From Heisenberg's uncertainity principle ,

\large {\sf {\underline{\bold {\boxed{\bigstar{ \: (\triangle x)(m \triangle v) = \dfrac{h}{4\pi} }}}}}}

★(△x)(m△v)=

4π

h

Where ,

Δx is uncertainity in position

m is mass

Δv is uncertainity in velocity

h is planck's constant

\begin{gathered} \sf: \implies \: ( \triangle x)[ (9.1 \times {10}^{ - 31} ( 3 \times {10}^{ - 3})]= \dfrac{6.625 \times {10}^{ - 34} }{4 \times 3.14} \\ \\ \sf: \implies (\triangle x)(27.3 \times {10}^{ - 34} ) = \dfrac{6.625 \times {10}^{ - 34} }{4 \times 3.14} \\ \\ \sf : \implies( \triangle x) = \dfrac{6.625 \times {10}^{ - 34} }{4 \times 3.14 \times 27.3 \times {10}^{ - 34} } \\ \\\sf : \implies (\triangle x) = \dfrac{6.625 \times \cancel{ {10}^{ - 34} }}{342.888 \times \cancel{ {10}^{ - 34} }} \\ \\ \sf : \implies ( \triangle x) = \dfrac{6.625}{342.888} \\ \\ \sf : \implies {\bold {\boxed {\pink{(\triangle x) = 0.019 \: m}}}}\end{gathered}

:⟹(△x)[(9.1×10

−31

(3×10

−3

)]=

4×3.14

6.625×10

−34

:⟹(△x)(27.3×10

−34

)=

4×3.14

6.625×10

−34

:⟹(△x)=

4×3.14×27.3×10

−34

6.625×10

−34

:⟹(△x)=

342.888×

10

−34

6.625×

10

−34

:⟹(△x)=

342.888

6.625

:⟹

(△x)=0.019m

∴ The uncertainity in position of the electron is 0.019 m