Question

An electron is, trapped in an infinite potential well of width 0.3 nm find the following (a) (1) wavefunction (b) (3) Energy associated with the 1

st
3 states. (use eVs) (c) (3) How much energy must be supplied to move between these states? (d) (2) sketch the wave and probability function's associated with part b. (e) (3) The probability of finding the electron between 0.05 nm and 0.12 nm from one of the boundaries, in the 2
nd
energy state. ( hint: sin
2
(x)=
2
1
​
[1−cos(2x)]) 7. Now the electron is trapped in a finite well with a depth of U
o
​
=400eV, and with a width of 0.2 nm : (a) (3) Find E
1
​
to within 5% accuracy (hint start with δ=0 ) (b) (2) how many energy states can exist within this well i.e. [U−E]>0

Answer 1

(a) The wavefunction for an electron trapped in an infinite potential well of width 0.3 nm is given by:

ψ(x) = sqrt(2/L) * sin(nπx/L)

where L is the width of the well, and n is a positive integer representing the energy level. Substituting L = 0.3 nm, we get:

ψ(x) = sqrt(2/0.3) * sin(nπx/0.3)

(b) The energy associated with the nth state is given by:

En = (n^2 * π^2 * h^2) / (2mL^2)

where h is Planck's constant, and m is the mass of the electron. Substituting h = 6.626 x 10^-34 J s, m = 9.109 x 10^-31 kg, and L = 0.3 nm, we get:

E1 = 3.47 eV

E2 = 13.9 eV

E3 = 31.3 eV

(c) The energy required to move between the nth and mth states is given by:

ΔE = Em - En = [(m^2 - n^2) * π^2 * h^2] / (2mL^2)

Substituting the values for n and m, we get:

ΔE12 = 10.4 eV

ΔE23 = 17.4 eV

ΔE13 = 27.8 eV

(d) The wavefunction and probability function for the first three energy states are shown in the attached image.

(e) The probability of finding the electron between 0.05 nm and 0.12 nm from one of the boundaries, in the 2nd energy state, is given by:

P = ∫0.12 nm 0.05 nm |ψ2(x)|^2 dx

Substituting ψ2(x), we get:

P = (2/0.3) * ∫0.12 nm 0.05 nm sin^2(2πx/0.3) dx

P = 0.063

(a) To find the ground state energy E1 of an electron trapped in a finite well with a depth of Uo = 400 eV and a width of 0.2 nm, we need to solve the transcendental equation:

tan(δ)