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Rzqust [24]
3 years ago
15

To form a hydrogen atom, a proton is fixed at a point and an electron is brought from far away to a distance of 0.529 × 10−10 m,

the average distance between proton and electron in a hydrogen atom. How much work is done?
Physics
2 answers:
WARRIOR [948]3 years ago
8 0

Given Information:

distance between electron and proton = d = 0.529x10⁻¹⁰ m

charge on electron = q₁ = -1.60x10⁻¹⁹ C

charge on proton = q₂ = 1.60x10⁻¹⁹ C

Required Information:

Work done = W = ?

Answer:

Work done = 4.35x10⁻¹⁸ Joules

Explanation:

We know that work done is given by

Work done = kq₁q₂/r

Where q₁ is the charge on electron, q₂ is the charge on proton and k is the coulomb constant and its value is 9x10⁹ Nm²/C²

Work done = (9x10⁹*-1.60x10⁻¹⁹*1.60x10⁻¹⁹)/0.529x10⁻¹⁰

Work done = 4.35x10⁻¹⁸ Joules

Therefore, 4.35x10⁻¹⁸ Joules of work is required to bring electron from a distance of 0.529x10⁻¹⁰ m towards the proton to form a hydrogen atom.

AlekseyPX3 years ago
6 0

Answer:

-4.37 * 10^(-18) J

Explanation:

Parameters given:

Charge of a proton, Qp = 1.6022 * 10^(-19) C

Charge of an electron, Qe = -1.6022 * 10^(-19) C

Distance between proton and electron, d = 0.529 × 10^(−10) m

The amount of work done is given as:

W = F * d

Where F = electrostatic force

d = distance between them

Electrostatic force is given as:

F = (k * Qp * Qe) / d²

F = (9 * 10^9 * 1.6022 * 10^(-19) * -1.6022 * 10^(-19)) / (0.529 × 10^(−10))²

F = -8.26 * 10^(-8) N

Work done will be:

W = -8.26 * 10^(-8) * 0.529 × 10^(−10)

W = -4.37 * 10^(-18) J

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A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
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5.82812 rad/s

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I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

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Answer:

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Weight lost in the 11 weeks = 11×2 =22 pounds

Total weight lost

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Tina has lost 24 pounds in total during the 13 weeks

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