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astraxan [27]
3 years ago
13

. Find the average distan e between the ele tron and proton in the Hydrogen ground-state. Use: hri = ˆ[infinity] 0 dr rP(r) wher

e ϕ(r) = Ae−r/aB is the ground-state solution of the radial S hrodinger equation for the Hydrogen atom (aB is Bohr's radius), and P(r) = 4πr2 |ϕ(r)| 2 is the probability density that the ele tron would be lo ated in the spheri al shell between r and r + dr. First normalize the wavefun tion to obtain the onstant A: ˆ[infinity] 0 dr P(r) = 1
Physics
1 answer:
CaHeK987 [17]3 years ago
7 0

Answer:

As data is given in the problem statement is below

ϕ(r) = Ae−r/aB

P(r) = 4πr2 |ϕ(r)| 2

r=0

After normalizing the distance between electron and proton will be 200 nm

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3 years ago
⦁ Match the following terms:
ki77a [65]

Answer:

Mass number - ⦁ The number of protons and neutrons in the nucleus of an atom.

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4 0
2 years ago
You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
2 years ago
Answer the following question​
Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0
2 years ago
A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict
Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

F=157\times 0.19\\\\F=29.83\ N

(B) The force is simply given by :

F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

6 0
3 years ago
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