A particle is in the ground state of a one-dimensional box of length 1.0 m. What is the minimum value of its momentum (in kg ⋅ m/s)? A) 9.9 × 10−³⁴ B) 6.6 × 10−³⁴ C) 3.3 × 10−³⁴ D) 13 × 10−³⁴ E) cannot be solved unless mass of particle is known.

Question

Quizplus · Accepted Answer

For a particle in a one-dimensional box of length L, the minimum momentum occurs at the first excited state with n = 2, given by: 
$p = \sqrt{\frac{2\hbar^2\pi^2}{mL^2}(n^2)}$
Substituting the known values, we get: 
$p = \sqrt{\frac{2\hbar^2\pi^2}{m(1	ext{ m})^2}(2^2)}$
$p = \sqrt{\frac{8\pi^2\hbar^2}{m}}$
Since we do not know the mass of the particle, we cannot calculate the exact value of the momentum. However, we can compare the values given in the answer choices. The closest value to the minimum of $\sqrt{8\pi^2\hbar^2/m}$ is option C: 3.3 × 10^-34 kg⋅m/s.

A Particle Is in the Ground State of a One-Dimensional