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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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The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an
blsea [12.9K]

Answer:

  g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

        F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

      F = G Me /Re²  m

We call gravity acceleration a

       g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

            R = Re + h

Therefore  the attractive force is

      F = G Me m / (Re + h)²

Let's take Re's common factor

      F = G Me / Re²  m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

         (1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

       F = G Me /Re²   m [1- 2 h / Re + 3 (h / Re)²]

       F = g₀   m  [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

     m g =g₀ m  [1- 2 h / Re + 3 (h / Re)²]

       g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

3 0
3 years ago
Which describes the changes in visible light moving from red to violet?
Alexxx [7]
The energy increases
5 0
2 years ago
A 60kg bicyclist (including the bicycle) is pedaling to the
Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight (W=mg): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force (F_A): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag (R): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

F_{net}=ma

where

F_{net} is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

a=3.1 m/s^2 is its acceleration

Substituting, we find the net force on the bicyclist:

F_{net}=(60)(3.1)=186 N

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

F_{net}=F_a-R

where:

F_{net} is the net force

F_a is the applied force (forward)

R is the air drag (backward)

In this problem we have:

F_{net}=186 N is the net force (found in part b)

R=60 N is the magnitude of the air drag

Solving for F_a, we find the force produced by the bicyclist while pedaling:

F_a=F_{net}+R=186+60=246 N

3 0
2 years ago
How does size of an object affect its gravity
irakobra [83]

Answer:

the more particles packed together the faster it falls

Explanation:

the mass + the 1 constant g-force = the speed without adding air resistance

6 0
2 years ago
A cow wanders 30 m North, turns 22 degrees right of its original path, and wanders another 40 m. Find its total displacement.
scoundrel [369]

Answer:OB=58.3m

Explanation:

So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.

now take the starting point as a origin such that cow moves in x-y co-ordinate axis.

As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.

So the final displacement is the length of cow from the origin that is length OB.

now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]

now displacement of cow= length of OB

                                           = \sqrt{[37.084]^{2}+[44.984]^{2}  }

                                           =\sqrt{3398.78}

                                     OB =58.3

                                         

7 0
9 months ago
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