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nalin [4]
3 years ago
7

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

uncertainty in the electron's momentum
Physics
1 answer:
Alina [70]3 years ago
4 0

Answer:

  • Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
  • Uncertainty in momentum (∆P) = ?
  • Planck's constant (h) = 6.26 × 10⁻³⁴ Js

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

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} {4 \times  \frac{22}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34}} { \frac{88}{7} }

\longrightarrow \:  \:  \sf24 \times  {10}^{ - 15}  .\Delta p =  \dfrac{6.26 \times  {10}^{ - 34} \times 7} { 8 }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 8  \times 24 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34} } { 192 \times  {10}^{ - 15} }

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ - 34}  \times  {10}^{15} } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{43.82 \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 2}  \times  {10}^{ -19}   } { 192}

\longrightarrow \:  \:  \sf\Delta p =  \dfrac{4382 \times  {10}^{ - 21}   } { 192}

\longrightarrow \:  \:  \sf\Delta p = 22.822\times  {10}^{ - 21}

\longrightarrow \:  \:  \sf\Delta p = 2.2822 \times  {10}^{1} \times  {10}^{ - 21}

\longrightarrow \:  \: \underline{ \boxed{ \red{  \bf\Delta p = 2.2822 \times  {10}^{ - 20}  \:  kg/ms}}}

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inserting the values in the above equation

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Answer: (a) Z-score are 1 and -1.2 for northern and southern regions, respectively.

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The z-score is calculated by the following:

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For the southern region:

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Answer:

ΔU = 5.21 × 10^(10) J

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We are given;

Mass of object; m = 1040 kg

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U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

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G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

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