Question

When the uncertainty in momentum is zero then the uncertainty in the position of a particle is

• A. \( \dfrac{h}{4\pi} \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( \dfrac{1}{2} \)
• E. infinity

Hint:

Remember the uncertainty relation: \[ \Delta x \Delta p \geq \frac{h}{4\pi} \] If one uncertainty becomes extremely small, the other must become extremely large.

Answer 1

Answer: (E)

Step 1: Recall Heisenberg’s uncertainty principle.
The uncertainty principle states that: \[ \Delta x \, \Delta p \geq \frac{h}{4\pi} \] where \[ \Delta x = \text{uncertainty in position} \] and \[ \Delta p = \text{uncertainty in momentum} \] This means position and momentum cannot both be exactly known at the same time.

Step 2: Use the given condition.
The question says that the uncertainty in momentum is zero: \[ \Delta p = 0 \] We substitute this into the uncertainty relation.

Step 3: Analyze the product \( \Delta x \Delta p \).
If \[ \Delta p = 0 \] then for the inequality \[ \Delta x \Delta p \geq \frac{h}{4\pi} \] to remain valid, \( \Delta x \) cannot be finite.
Because any finite number multiplied by \( 0 \) would give \( 0 \), which cannot be greater than or equal to \[ \frac{h}{4\pi} \]

Step 4: Determine what \( \Delta x \) must be.
The only way to interpret this physically is: \[ \Delta x \to \infty \] That is, if momentum is known exactly, the position becomes completely uncertain.

Step 5: Understand the physical meaning.
Exact momentum means the particle behaves like an ideal plane wave.
A plane wave is spread over all space, so its position cannot be localized.
Hence the uncertainty in position becomes infinite.

Step 6: Match with the options.
Among the given choices, the correct one is: \[ \text{infinity} \]

Step 7: Final conclusion.
Therefore, when \[ \Delta p = 0 \] the uncertainty in position is \[ \boxed{\infty} \] Hence, the correct option is \[ \boxed{(5)\ \text{infinity}} \]