A particle with the mass of a proton is trapped in a box. The particle falls from the n=3 state to the n=2 state and emits 0.16 meV of energy. What is the size of the box? Please report your answer in nm.
To find the size of the box, we can use the energy difference between the two states and apply the formula for the energy of a confined particle in a box:
E = (n^2 * h^2) / (8 * m * L^2)
Where:
E is the energy difference between the states,
n is the quantum number,
h is the Planck's constant (6.626 x 10^-34 Js),
m is the mass of the particle,
L is the size of the box.
We are given:
- E = 0.16 meV
- n1 = 3
- n2 = 2
- m (mass of proton) = 1.67 x 10^-27 kg
First, we convert the given energy from millielectron volts (meV) to joules (J) using the conversion factor: 1 meV = 1.6 x 10^-22 J.
E = (0.16 meV) * (1.6 x 10^-22 J/meV)
E = 2.56 x 10^-23 J
Now, we substitute the values into the formula and solve for L:
2.56 x 10^-23 J = [(3^2 * (6.626 x 10^-34 Js)^2) / (8 * (1.67 x 10^-27 kg) * L^2)]
Simplifying the equation further:
2.56 x 10^-23 J = (9 * (6.626 x 10^-34 Js)^2) / (8 * (1.67 x 10^-27 kg) * L^2)
Multiplying both sides by L^2 and rearranging the equation:
(9 * (6.626 x 10^-34 Js)^2) / (8 * (1.67 x 10^-27 kg) * (2.56 x 10^-23 J)) = L^2
Taking the square root of both sides to find L:
L = sqrt((9 * (6.626 x 10^-34 Js)^2) / (8 * (1.67 x 10^-27 kg) * (2.56 x 10^-23 J)))
Calculating the value:
L ≈ 2.285 x 10^-10 m ≈ 228.5 pm
Therefore, the size of the box is approximately 228.5 picometers (pm), which is equivalent to 0.2285 nanometers (nm).