Question

an electron confined in a one-dimensional box is observed, at different times, to have energies of 12 evev, 27 evev, and 48 evev. part a what is the minimal length of the box? express your answer in nanometers.

Answer 1

The minimal length of the box for the given energies of 12 eV, 27 eV, and 48 eV is approximately 9.874 nm.

Step-by-step explanation:

The minimal length of the box can be calculated by using the formula:

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

Where:

• L is the length of the box
• n is the energy level of the electron
• h is Planck's constant (6.62607015 × 10^-34 J·s)
• m is the mass of the electron (9.10938356 × 10^-31 kg)
• E is the energy of the electron

For the given energies of 12 eV, 27 eV, and 48 eV, we can calculate the corresponding lengths of the box. Plugging the values into the formula, we get:

L = (2 * n^2 * (6.62607015 × 10^-34 J·s)^2) / (8 * (9.10938356 × 10^-31 kg) * E)

L = (2 * 1^2 * (6.62607015 × 10^-34 J·s)^2) / (8 * (9.10938356 × 10^-31 kg) * 12 eV)

L = 23.461 nm (approx)

L = (2 * 2^2 * (6.62607015 × 10^-34 J·s)^2) / (8 * (9.10938356 × 10^-31 kg) * 27 eV)

L = 14.097 nm (approx)

L = (2 * 3^2 * (6.62607015 × 10^-34 J·s)^2) / (8 * (9.10938356 × 10^-31 kg) * 48 eV)

L = 9.874 nm (approx)

Therefore, the minimal length of the box is approximately 9.874 nm.