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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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The answer is A ..........
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What is the most important factor for the formation of our planets
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The gravitational pull of the Sun the interstellar dust attracting heat away from the protosun the process of nuclear fusion the nebular cloud condensing.
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3 years ago
After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T
motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

KE_r = \dfrac{1}{2}I\omega^2

I of the moment of inertia of the sphere

I = \dfrac{2}{5}MR^2

 v = R ω

using conservation of energy

KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2

KE_r = \dfrac{1}{5}MV^2

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h

0.7 V^2 = 0.7 V'^2 + gh

0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0
3 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
Westkost [7]

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

5 0
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