Question

(a) Make a qualitative sketch of the wave function for the \(n=5\) state of an electron in a finite box \([U(x)=0\) for \(00\) elsewherel. (b) If \(L=1.0 \mathrm{nm}\) and \(U_{0}=1.0\) keV, estimate the number of bound states that exist.

Short answer

Answer: The qualitative sketch of the wave function for the n=5 state of an electron in a finite box consists of a sine curve passing through 4 equidistant nodes within the box and exponential functions outside the box. For the given parameters L = 1.0 nm and U0 = 1000 eV, there are 4 bound states.

Step-by-step solution

Part (a): Draw a qualitative sketch for the wave function of the n=5 state

To draw the qualitative sketch of the wave function for the n=5 state, we need to understand the behavior of the wave function inside and outside the box. Inside the box, the wave function follows the equation: Ψ(x) = A sin(kx) + B cos(kx) Where k = (2πn) / L, and A and B are constants. Outside the box, the wave function behaves exponentially, as described by Ψ(x) = Ce^(-λx) for x > L Ψ(x) = De^(-λx) for x < 0 Where λ = sqrt(2mU0)/ħ and C and D are constants. For the n=5 state, the wave function inside the box will have 4 nodes (points where the wave function touches zero) and follow a sine curve: Ψ(x) = A sin((2π * 5) / L * x) Now, we can sketch the wave function for the n=5 state: 1. Draw 4 equidistant nodes within the box (between 0 and L) 2. Sketch a sine curve that passes through these nodes within the box 3. Sketch the exponential functions outside the box (x < 0 and x > L)

Part (b): Estimate the number of bound states for the given parameters L and U0

To estimate the number of bound states, we will use the following equation: E_n = [(n^2 * h^2) / (8 * m * L^2)] + U0 Where E_n is the energy of the nth bound state, n is an integer, h is the Planck's constant, m is the mass of the electron, and L is the width of the box. From this equation, we can see that the bound states exist when E_n < U0. Let's now substitute the values L = 1.0 nm and U0 = 1000 eV (1 keV) into the equation: E_n = [(n^2 * (6.626 * 10^-34 J s)^2) / (8 * (9.11 * 10^-31 kg) * (1.0 * 10^-9 m)^2)] + 1000 eV Divide both sides by 1.6*10^(-19) to convert energy to eV: E_n = [(n^2 * (6.626 * 10^-34 J s)^2) / (8 * (9.11 * 10^-31 kg) * (1.0 * 10^-9 m)^2 * 1.6 * 10^(-19) J/eV)] + 1000 eV To find the maximum value of n that satisfies E_n < U0, plug in the known values and solve for n: E_n < 1000 n^2 < [(1000 - [(6.626 * 10^-34 J s)^2 / (8 * (9.11 * 10^-31 kg) * (1.0 * 10^-9 m)^2 * 1.6 * 10^(-19) J/eV)]) * (8 * (9.11 * 10^-31 kg) * (1.0 * 10^-9 m)^2) / (((6.626 * 10^-34 J s)^2))] Solve for n: n < sqrt( [(1000 - [(6.626 * 10^-34 J s)^2 / (8 * (9.11 * 10^-31 kg) * (1.0 * 10^-9 m)^2 * 1.6 * 10^(-19) J/eV)]) * (8 * (9.11 * 10^-31 kg) * (1.0 * 10^-9 m)^2) / (((6.626 * 10^-34 J s)^2))]) n < 4.35 The integer value less than 4.35 is 4. Therefore, there are 4 bound states for the given parameters L and U0.