Question

A free particle of mass $m$ confined within the walls of a one-dimensional box of length $a$ (with the potential outside the box being infinity), is in the eigenstate having quantum number $n = 4$The magnitude of uncertainty in the measurement of momentum of the particle is $Y \times \frac{h}{a}$The value of $Y$ is _ _ _. (answer in integer)

Hint:

For particle in a box, $\Delta p = \frac{nh}{2a}$ and $\langle p \rangle = 0$Always use variance definition for uncertainty

Answer 1

Correct Answer: 2

Step 1: Write momentum eigenvalues in a 1D box.
For a particle in a box, momentum is not sharply definedThe expectation value of momentum is zero but uncertainty exists
Allowed momentum magnitude:
\[ p_n = \frac{nh}{2a} \]

Step 2: Recall uncertainty definition.
\[ (\Delta p)^2 = \langle p^2 \rangle - \langle p \rangle^2 \]
For stationary states:
\[ \langle p \rangle = 0 \]
Thus,
\[ \Delta p = \sqrt{\langle p^2 \rangle} \]

Step 3: Calculate $\langle p^2 \rangle$.
For particle in box:
\[ \langle p^2 \rangle = \left(\frac{nh}{2a}\right)^2 \]

Step 4: Find uncertainty in momentum.
\[ \Delta p = \frac{nh}{2a} \]

Step 5: Substitute $n = 4$.
\[ \Delta p = \frac{4h}{2a} = \frac{2h}{a} \]

Step 6: Express in given form.
\[ \Delta p = Y \times \frac{h}{a} \]
\[ Y = 2 \]

Step 7: Correction using standard quantum result.
Exact uncertainty for particle in box:
\[ \Delta p = \frac{nh}{2a} \Rightarrow Y = \frac{n}{2} = 2 \]
Thus,
\[ \boxed{2} \]