Question

A particle with a mass of 1.5 x 10^(-27) kg is confined to a one-dimensional potential energy well of width 2 nm.

a) Determine the ground state energy of the particle in electron volts (eV).

b) Calculate the uncertainty in momentum of the particle in kilograms meters per second (kg·m/s) due to the Heisenberg uncertainty principle.

c) What is the wavelength associated with this uncertainty in momentum in nanometers (nm)?

Answer

a) To determine the ground state energy of the particle, we can use the formula for the quantized energy levels in an infinite square well potential:

E_n = (n^2 * h^2) / (8 * m * L^2)

Where:

• E_n is the energy of the nth energy level
• n is the quantum number (n = 1 represents the ground state)
• h is the Planck's constant (6.626 x 10^(-34) J·s)
• m is the mass of the particle
• L is the width of the potential energy well

Converting the given mass to kilograms:

m = 1.5 x 10^(-27) kg

Substituting the values into the formula:

E_1 = (1^2 * (6.626 x 10^(-34) J·s)^2) / (8 * (1.5 x 10^(-27) kg) * (2 x 10^(-9) m)^2)

Simplifying the equation:

E_1 = (1 * (6.626 x 10^(-34))^2) / (8 * 1.5 x 10^(-27) * 2^2 * 10^(-18))

E_1 = 6.426 x 10^(-12) J

Converting the energy to electron volts (1 eV = 1.6 x 10^(-19) J):

E_1 = (6.426 x 10^(-12) J) / (1.6 x 10^(-19) J/eV)

E_1 = 40.15 eV

Therefore, the ground state energy of the particle is 40.15 eV.

b) The Heisenberg uncertainty principle states that the uncertainty in momentum (Δp) and position (Δx) of a particle are related by the equation:

Δp * Δx >= h / (4π)

Where:

• Δp is the uncertainty in momentum
• Δx is the uncertainty in position
• h is the Planck's constant (6.626 x 10^(-34) J·s)
• π is a mathematical constant

To calculate the uncertainty in momentum, we rearrange the equation:

Δp >= (h / (4π * Δx))

Substituting the width of the potential energy well as the uncertainty in position:

Δx = 2 nm = 2 x 10^(-9) m

Δp >= (6.626 x 10^(-34) J·s) / (4π * 2 x 10^(-9) m)

Simplifying the equation:

Δp >= 2.101 x 10^(-25) kg·m/s

Therefore, the uncertainty in momentum of the particle is 2.101 x 10^(-25) kg·m/s.

c) The wavelength associated with the uncertainty in momentum can be found using the de Broglie wavelength equation:

λ = h / p

Where:

• λ is the wavelength
• h is the Planck's constant (6.626 x 10^(-34) J·s)
• p is the momentum of the particle

Substituting the calculated uncertainty in momentum:

λ = (6.626 x 10^(-34) J·s) / (2.101 x 10^(-25) kg·m/s)

Simplifying the equation:

λ = 3.153 x 10^(-9) m

Converting the wavelength to nanometers (1 nm = 10^(-9) m):

λ = 3.153 x 10^(-9) m * 10^9 nm/m

λ = 3.153 nm

Therefore, the wavelength associated with the uncertainty in momentum is 3.153 nm.