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Slav-nsk [51]
4 years ago
13

The nuclear potential that binds protons and neutrons in the nucleus of an atom is often approximated by a square well. Imagine

a proton conned in an innite square well of length 105 nm, a typical nuclear diameter. Calculate the wavelength and energy associated with the photon that is emitted when the proton undergoes a transition from the rst excited state (n 2) to the ground state (n 1). In what region of the electromagnetic spectrum does this wavelength belong?
Physics
1 answer:
slega [8]4 years ago
5 0

3. The nuclear potential that binds protons and neutrons in the nucleus of an atom

is often approximated by a square well. Imagine a proton confined in an infinite

square well of length 10−5 nm, a typical nuclear diameter. Calculate the wavelength

and energy associated with the photon that is emitted when the proton undergoes a

transition from the first excited state (n = 2) to the ground state (n = 1). In what

region of the electromagnetic spectrum does this wavelength belong?

Answer 3

We are given that,

Length of square well = L = 10−5

nm = 10−14 m.

Energy of proton in state n is given by,

En =

π

2n

2~

2

2mpL2

,

where L is the width of the square well.

⇒ E1 =

π

2~

2

2mpL2

E2 =

4π

2~

2

2mpL2

·

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disa [49]

Answer:

4.15 m/s

Explanation:

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ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = \frac{1}{2} *m*vf^{2}  (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

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So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0
4 years ago
A proton travels with a speed of 1.8×106 m/s at an angle 53◦ with a magnetic field of 0.49 T pointed in the +y direction. The ma
Likurg_2 [28]

Answer:

Magnetic force, F=1.12\times 10^{-13}\ N

Explanation:

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The magnitude of magnetic force is given by :

F=qvB\ sin\theta

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F=1.12\times 10^{-13}\ N

So, the magnitude of the magnetic force on the proton is 1.12\times 10^{-13}\ N. Hence, this is the required solution.

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kipiarov [429]
3 impaired metabolic processes
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How is lithium fluoride and lithium chloride similar?
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Answer:

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Explanation:

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The wavelength of the first harmonic of the standing wave is 2L.

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We know from  the formula of the first harmonic that the wavelength of the first harmonic of the standing wave is 2L.

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