Question

a) Calculate the components of energy along x, y and z axes and the total energy for an electron in a cubical box of length 10^-9 m, if nx = 3, ny =nz =1. State the values of nx, ny and nz for two other energy states which are degenerate with this level.

Hint: Use the principle of calculation of energy of a particle in the three dimensional box. (5)

Answer 1

Answer on Question #44189, Physics, Quantum Mechanics

Question:

a) Calculate the components of energy along $x, y$ and $z$ axes and the total energy for an electron in a cubical box of length $10^{\wedge} - 9\,\mathrm{m}$, if $nx = 3$, $ny = nz = 1$. State the values of $nx$, $ny$ and $nz$ for two other energy states which are degenerate with this level.

Hint: Use the principle of calculation of energy of a particle in the three-dimensional box. (5)

Answer:

The time-independent Schrodinger equation:

$$- \frac {\hbar^ {2}}{2 m} \nabla^ {2} \psi (r) + V (r) \psi (r) = E \psi (r)$$

Since we are dealing with a 3-dimensional figure, we need to add the 3 different axes into the Schrodinger equation:

$$- \frac {\hbar^ {2}}{2 m} \left(\frac {d ^ {2} \psi (r)}{d x ^ {2}} + \frac {d ^ {2} \psi (r)}{d y ^ {2}} + \frac {d ^ {2} \psi (r)}{d z ^ {2}}\right) = E \psi (r)$$

The easiest way in solving this partial differential equation is by having the wave function equal to each individual function for its individual variable:

$$\Psi (x, y, z) = X (x) Y (y) Z (z)$$

Now each function has its own variable:

Now substitute $\Psi(x, y, z)$ into Schrodinger equation and divide it by the product, $X(x)Y(y)Z(z)$:

$$- \frac {\hbar^ {2}}{2 m} \left(\frac {1}{X (x)} \frac {d ^ {2} \psi (r)}{d x ^ {2}} + \frac {1}{Y (y)} \frac {d ^ {2} \psi (r)}{d y ^ {2}} + \frac {1}{Z (z)} \frac {d ^ {2} \psi (r)}{d z ^ {2}}\right) = E$$

Now separate each term in last equation to equal zero:

$$\frac {d ^ {2} X}{d x ^ {2}} + \frac {2 m}{\hbar^ {2}} E _ {x} X = 0$$ $$\frac {d ^ {2} Y}{d x ^ {2}} + \frac {2 m}{\hbar^ {2}} E _ {y} X = 0$$ $$\frac {d ^ {2} Z}{d x ^ {2}} + \frac {2 m}{\hbar^ {2}} E _ {z} X = 0$$

Solution for this equation is (the same for variables y and z):

$$E _ {x} = \frac {n _ {x} ^ {2} h ^ {2}}{8 m a ^ {2}}$$

Now we can add all the energies together to get the total energy:

$$E = E _ {x} + E _ {y} + E _ {z} = \frac {h ^ {2}}{8 m a ^ {2}} \left(n _ {x} ^ {2} + n _ {y} ^ {2} + n _ {z} ^ {2}\right)$$

The component of energy along $x$ axis:

$$E _ {x} = \frac {n _ {x} ^ {2} h ^ {2}}{8 m a ^ {2}} \cong 3.39 \, \text{eV}$$

The components of energy along y and z axes:

$$E _ {y} = E _ {z} = \frac {n _ {x} ^ {2} h ^ {2}}{8 m a ^ {2}} = \frac {h ^ {2}}{8 m a ^ {2}} \cong 0.38 \, \text{eV}$$

Total energy equals:

$$E = E _ {x} + E _ {y} + E _ {z} \cong 4.15 \, \text{eV}$$

Other 2 states with the same energy are:

$$n _ {x} ^ {2} + n _ {y} ^ {2} + n _ {z} ^ {2} = 3 ^ {2} + 1 + 1 = 11$$ $$n _ {x} = 1, n _ {y} = 1, n _ {z} = 3$$ $$n _ {x} = 1, n _ {y} = 3, n _ {z} = 1$$

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