Question

A golf ball has a mass of 100g, and a speed of 10 m/s. If the speed can be measured within accuracy
of 2%, calculate the uncertainty in the position(metre)
1)10-30
2)331.5x10-34
3)10-28
4)6.63x10-31​

Answer 1

Answer:

Heisenbergs uncertainty Principle:

• mass of golf ball (m) = 100 g = 100/1000 = 0.1 kg

• Exact value = 10 m/s

• Accuracy = 2%

• Uncertainty in velocity (∆v) = ?

• Uncertainty in position (∆x) = ?

First of all let's calculate the uncertainty in velocity (∆v) :

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

$\longrightarrow \: \sf \Delta v = \dfrac{10 \times 2}{100}$

$\longrightarrow \: \sf \Delta v = \dfrac{20}{100}$

$\longrightarrow \: \sf \Delta v = \dfrac{2}{10}$

$\longrightarrow \: \sf \Delta v = \dfrac{1}{5}$

$\longrightarrow \: \sf \Delta v = 0.2$

Now, we can find the uncertainty in position (∆x) :

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

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

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

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

$\longrightarrow \: \sf \Delta x \times 0.2= \dfrac{0.53 \times {10}^{ - 34} }{0.1}$

$\longrightarrow \: \sf \Delta x = \dfrac{0.53 \times {10}^{ - 34} }{0.1 \times 0.2}$

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

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

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

$\longrightarrow \: \sf \Delta x = \dfrac{0.53 \times {10}^{ - 32} }{ 2 }$

$\longrightarrow \: \sf \Delta x = 0.265\times {10}^{ - 32}$

$\longrightarrow \: \sf \Delta x = 2.65\times {10}^{ - 1} \times {10}^{ - 32}$

$\longrightarrow \: \underline{ \boxed{ \orange{\bf \Delta x = 2.65\times {10}^{ - 33} \: m}}}$