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Likurg_2 [28]
2 years ago
12

The amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is________

__. Let the radius (r) of atom is 52 pm a) Zero b) infinity c) Fr d) 2Fr
Physics
1 answer:
maksim [4K]2 years ago
6 0

Answer:

d) 2Fr

Explanation:

We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell  -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².

We now evaluate the integral from r = +r to -r

W = ∫Fdr

= ∫(-e²/4πε₀r²)dr

= -∫e²dr/4πε₀r²

= -e²/4πε₀∫dr/r²

= -e²/4πε₀ × -[1/r] from r = +r to -r

W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.

Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.

So W = -2e²/4πε₀r = 2Fr.

So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr

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MAVERICK [17]
Protons do not move out of the nucleus of atoms although they repel each other.

Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.
8 0
2 years ago
An object takes 5 seconds to move 2 meters upward. How fast did it go?
jonny [76]

Answer:

2.5

Explanation:

5/2=2.5

8 0
2 years ago
A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th
Gala2k [10]

Answer:

F_a_v_g=7093333.33N*s

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}

Where:

m=mass\hspace{3}of\hspace{3}the\hspace{3}object

v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval

v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.

t_2=final\hspace{3}time

t_1=initial\hspace{3}time

Asumming v1=0 and t1=0:

F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s

8 0
3 years ago
A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find
vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, a_c = v²/r

∴ The centripetal acceleration, a_c = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, a_c = 20·π/9 m/s² ≈ 6.98 m/s²

4 0
2 years ago
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
denpristay [2]

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

3 0
3 years ago
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