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2 2 m d 2 d x 2 ( B k sin ( k x ) ) = E ( B k sin ( k x ) ) .

Computing these derivatives leads to

E = E k = 2 k 2 2 m .

According to de Broglie, p = k , so this expression implies that the total energy is equal to the kinetic energy, consistent with our assumption that the “particle moves freely.” Combining the results of [link] and [link] gives

E n = n 2 π 2 2 2 m L 2 , n = 1 , 2 , 3 , . . .

Strange! A particle bound to a one-dimensional box can only have certain discrete (quantized) values of energy. Further, the particle cannot have a zero kinetic energy—it is impossible for a particle bound to a box to be “at rest.”

To evaluate the allowed wave functions that correspond to these energies, we must find the normalization constant B n . We impose the normalization condition [link] on the wave function

ψ n ( x ) = B n sin n π x / L
1 = 0 L d x | ψ n ( x ) | 2 = 0 L d x B n 2 sin 2 n π L x = B n 2 0 L d x sin 2 n π L x = B n 2 L 2 B n = 2 L .

Hence, the wave functions that correspond to the energy values given in [link] are

ψ n ( x ) = 2 L sin n π x L , n = 1 , 2 , 3 , . . .

For the lowest energy state or ground state energy    , we have

E 1 = π 2 2 2 m L 2 , ψ 1 ( x ) = 2 L sin ( π x L ) .

All other energy states can be expressed as

E n = n 2 E 1 , ψ n ( x ) = 2 L sin ( n π x L ) .

The index n is called the energy quantum number    or principal quantum number    . The state for n = 2 is the first excited state, the state for n = 3 is the second excited state, and so on. The first three quantum states (for n = 1 , 2 , and 3 ) of a particle in a box are shown in [link] .

The wave functions in [link] are sometimes referred to as the “states of definite energy.” Particles in these states are said to occupy energy levels    , which are represented by the horizontal lines in [link] . Energy levels are analogous to rungs of a ladder that the particle can “climb” as it gains or loses energy.

The wave functions in [link] are also called stationary state     s and standing wave state     s . These functions are “stationary,” because their probability density functions, | Ψ ( x , t ) | 2 , do not vary in time, and “standing waves” because their real and imaginary parts oscillate up and down like a standing wave—like a rope waving between two children on a playground. Stationary states are states of definite energy [ [link] ], but linear combinations of these states, such as ψ ( x ) = a ψ 1 + b ψ 2 (also solutions to Schrӧdinger’s equation) are states of mixed energy.

The first three quantum states of a quantum particle in a box for principal quantum numbers n=1, n=2, and n=3 are shown: Figure (a) shown the graphs of the standing wave solutions. The vertical axis is the wave function, with a separate origin for each state that is aligned with the energy scale of figure (b). The horizontal axis is x from just below 0 to just past L. Figure (b) shows the energy of each of the states on the vertical E sub n axis. All of the wave functions are zero for x less than 0 and x greater than L. The n=1 function is the first half wave of the wavelength 2 L sine function and its energy is pi squared times h squared divided by the quantity 2 m L squared. The n=2 function is the first full wave of the wavelength 2 L sine function and its energy is 4 pi squared times h squared divided by the quantity 2 m L squared. The n=3 function is the first one and a half waves of the wavelength 2 L sine function and its energy is 9 pi squared times h squared divided by the quantity 2 m L squared.
The first three quantum states of a quantum particle in a box for principal quantum numbers n = 1 , 2 , and 3 : (a) standing wave solutions and (b) allowed energy states.

Energy quantization is a consequence of the boundary conditions. If the particle is not confined to a box but wanders freely, the allowed energies are continuous. However, in this case, only certain energies ( E 1 , 4 E 1 , 9 E 1 , …) are allowed. The energy difference between adjacent energy levels is given by

Δ E n + 1 , n = E n + 1 E n = ( n + 1 ) 2 E 1 n 2 E 1 = ( 2 n + 1 ) E 1 .

Conservation of energy demands that if the energy of the system changes, the energy difference is carried in some other form of energy. For the special case of a charged particle confined to a small volume (for example, in an atom), energy changes are often carried away by photons. The frequencies of the emitted photons give us information about the energy differences (spacings) of the system and the volume of containment—the size of the “box” [see [link] ].

Questions & Answers

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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.
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Sonu Reply
Show that n1Sino1=n2Sino2
javan Reply
what's propagation
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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?
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Practice Key Terms 7

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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