<< Chapter < Page Chapter >> Page >

A simple model of the nucleus

Suppose a proton is confined to a box of width L = 1.00 × 10 −14 m (a typical nuclear radius). What are the energies of the ground and the first excited states? If the proton makes a transition from the first excited state to the ground state, what are the energy and the frequency of the emitted photon?

Strategy

If we assume that the proton confined in the nucleus can be modeled as a quantum particle in a box, all we need to do is to use [link] to find its energies E 1 and E 2 . The mass of a proton is m = 1.76 × 10 −27 kg. The emitted photon carries away the energy difference Δ E = E 2 E 1 . We can use the relation E f = h f to find its frequency f .

Solution

The ground state:

E 1 = π 2 2 2 m L 2 = π 2 ( 1.05 × 10 −34 J · s ) 2 2 ( 1.67 × 10 −27 kg ) ( 1.00 × 10 −14 m ) 2 = 3.28 × 10 −13 J = 2.05 MeV .

The first excited state: E 2 = 2 2 E 1 = 4 ( 2.05 MeV ) = 8.20 MeV .

The energy of the emitted photon is E f = Δ E = E 2 E 1 = 8.20 MeV 2.05 MeV = 6.15 MeV .

The frequency of the emitted photon is

f = E f h = 6.15 MeV 4.14 × 10 −21 MeV · s = 1.49 × 10 21 Hz .

Significance

This is the typical frequency of a gamma ray emitted by a nucleus. The energy of this photon is about 10 million times greater than that of a visible light photon.

Got questions? Get instant answers now!

The expectation value of the position for a particle in a box is given by

x = 0 L d x ψ n * ( x ) x ψ n ( x ) = 0 L d x x | ψ n * ( x ) | 2 = 0 L d x x 2 L sin 2 n π x L = L 2 .

We can also find the expectation value of the momentum or average momentum of a large number of particles in a given state:

p = 0 L d x ψ n * ( x ) [ i d d x ψ n ( x ) ] = i 0 L d x 2 L sin n π x L [ d d x 2 L sin n π x L ] = i 2 L 0 L d x sin n π x L [ n π L cos n π x L ] = i 2 n π L 2 0 L d x 1 2 sin 2 n π x L = i n π L 2 L 2 n π 0 2 π n d φ sin φ = i 2 L · 0 = 0.

Thus, for a particle in a state of definite energy, the average position is in the middle of the box and the average momentum of the particle is zero—as it would also be for a classical particle. Note that while the minimum energy of a classical particle can be zero (the particle can be at rest in the middle of the box), the minimum energy of a quantum particle is nonzero and given by [link] . The average particle energy in the nth quantum state—its expectation value of energy—is

E n = E = n 2 π 2 2 2 m .

The result is not surprising because the standing wave state is a state of definite energy. Any energy measurement of this system must return a value equal to one of these allowed energies.

Our analysis of the quantum particle in a box would not be complete without discussing Bohr’s correspondence principle. This principle states that for large quantum numbers, the laws of quantum physics must give identical results as the laws of classical physics. To illustrate how this principle works for a quantum particle in a box, we plot the probability density distribution

| ψ n ( x ) | 2 = 2 L sin 2 ( n π x / L )

for finding the particle around location x between the walls when the particle is in quantum state ψ n . [link] shows these probability distributions for the ground state, for the first excited state, and for a highly excited state that corresponds to a large quantum number. We see from these plots that when a quantum particle is in the ground state, it is most likely to be found around the middle of the box, where the probability distribution has the largest value. This is not so when the particle is in the first excited state because now the probability distribution has the zero value in the middle of the box, so there is no chance of finding the particle there. When a quantum particle is in the first excited state, the probability distribution has two maxima, and the best chance of finding the particle is at positions close to the locations of these maxima. This quantum picture is unlike the classical picture.

Questions & Answers

Diagram of the derive rotational analog equation of v= u+at
Nnamnso Reply
what is carat
Arnulfo Reply
what is chemistry
Mrs Reply
what chemistry ?
Abakar
where are the mcq
Fred Reply
ok
Giorgi
acids and bases
Navya
How does unpolarized light have electric vector randomly oriented in all directions.
Tanishq Reply
unpolarized light refers to a wave collection which has an equal distribution of electric field orientations for all directions
pro
In a grating, the angle of diffraction for second order maximum is 30°.When light of wavelength 5*10^-10cm is used. Calculate the number of lines per cm of the grating.
Micheal Reply
state the law of gravity 6
cletus Reply
what is cathodic protection
Ebe Reply
its just a technique used for the protection of a metal from corrosion by making it cathode of an electrochemical cell.
akif
what is interferometer
Sonu Reply
what is interferometer
Abakar
Show that n1Sino1=n2Sino2
javan Reply
what's propagation
Vikas Reply
is it in context of waves?
Edgar
It is the manner of motion of the energy whether mechanical(requiring elastic medium)or electromagnetic(non interference with medium)
Edgar
determine displacement cat any time t for a body of mass 2kg under a time varrying force ft=bt³+csinkt
Felix Reply
A round diaphragm S with diameter of d = 0.05 is used as light source in Michelson interferometer shown on the picture. The diaphragm is illuminated by parallel beam of monochromatic light with wavelength of λ = 0.6 μm. The distances are A B = 30, A C = 10 . The interference picture is in the form of concentric circles and is observed on the screen placed in the focal plane of the lens. Estimate the number of interference rings m observed near the main diffractive maximum.
Jyoti Reply
A Pb wire wound in a tight solenoid of diameter of 4.0 mm is cooled to a temperature of 5.0 K. The wire is connected in series with a 50-Ωresistor and a variable source of emf. As the emf is increased, what value does it have when the superconductivity of the wire is destroyed?
Rupal Reply
how does colour appear in thin films
Nwjwr Reply
hii
Sonu
hao
Naorem
hello
Naorem
hiiiiii
ram
🎓📖
Deepika
yaaa ☺
Deepika
ok
Naorem
hii
PALAK
Practice Key Terms 7

Get the best University physics vol... course in your pocket!





Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 3' conversation and receive update notifications?

Ask