Question

A golf ball has a mass of 40g ,and a speed of 45m /s. If the speed can be measured within accuracy of 2%,calculate the uncertainty in the position. ​

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Answer 1

Answer:

• mass of ball (m) = 40 g = 40/1000 = 0.04 kg

• Accuracy = 2%

• Exact value = 45 m/s

• Planck's constant (h) = 6.63 × 10-³⁴ Js

Uncertainty in velocity:

$\longrightarrow\:\sf \Delta v = Exact \: value\times Accuracy$

$\longrightarrow\:\sf \Delta v = \dfrac{45 \times 2}{100}$

$\longrightarrow\:\sf \Delta v = \dfrac{90}{100}$

$\longrightarrow\:\sf \Delta v = \dfrac{9}{10}$

$\longrightarrow\: \underline{ \boxed{ \orange{\bf \Delta v = 0.9 \: {ms}^{ - 1}}}}$

Heisenbergs uncertainty Principle:

$\longrightarrow\:\sf \Delta x.\Delta v = \dfrac{h}{4\pi m}$

$\longrightarrow\:\sf \Delta x. 0.9= \dfrac{6.63 \times {10}^{ - 34} }{4 \times 3.14 \times0.04 }$

$\longrightarrow\:\sf \Delta x. 0.9= \dfrac{0.53 \times {10}^{ - 34} }{0.04 }$

$\longrightarrow\:\sf \Delta x= \dfrac{0.53 \times {10}^{ - 34} }{0.04 \times 0.9}$

$\longrightarrow\:\sf \Delta x= \dfrac{0.53 \times {10}^{ - 34} }{4 \times {10}^{ - 2} \times 9 \times {10}^{ - 1} }$

$\longrightarrow\:\sf \Delta x= \dfrac{0.53 \times {10}^{ - 34} }{ 36 \times {10}^{ -3} }$

$\longrightarrow\:\sf \Delta x= \dfrac{0.53 \times {10}^{ - 34} \times {10}^{3} }{ 36 }$

$\longrightarrow\:\sf \Delta x= \dfrac{53 \times {10}^{ - 31} \times {10}^{ - 2} }{ 36 }$

$\longrightarrow\: \underline{ \boxed{ \red{\bf \Delta x= 1.473 \times {10}^{ - 33 } \: m }}}$

Answer 2

Answer:

Use Heisenberg's uncertainty principle.

For example [Δx=

4πmΔv

h

]

Here, Δx is the uncertainty in the position.

Δv is the uncertainty in velocity

m is the mass of Particle.

Given,m=40g=0.04kg

Δv=2%ofv=2×

100

45

=0.9m/s

h=6.626×10

−34

J.s

Now,Δx=

(4×3.14×0.04×0.9)

6.626×10

−34

=1.4654×10

−33

m

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