If uncertainty in the position of the electron is zero, then its uncer

1.	If uncertainty in the position of the electron is zero, then its uncertainty in momentum will be:1. h / 2π2. h / 4π3. zero4. infinity
Answer» Correct Answer - Option 4 : infinity Concept: Heisenberg’s Uncertainty Principle: It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron. Mathematically, it can be given as \({\rm{\Delta }}x \times {\rm{\Delta }}{p_x} \ge \frac{h}{{4\pi }}\) ∆x is uncertaintyin position,∆px is uncertaintyin momentum at position x. Explanation: Now, it is given thatuncertainty in momentum is zero. ∆px = 0 But, by the uncertainty principle \({\rm{\Delta }}x \times {\rm{\Delta }}{p_x} \ge \frac{h}{{4\pi }}\) \(\implies {0} \times {\rm{\Delta }}{x} \ge \frac{h}{{4\pi }}\) \(\implies {\rm{\Delta }}{x} \ge \frac{h}{{4\pi \times 0 }} \) Now anything divided by zero is infinity. So,∆x is infinity. So, infinity is the correct answer.

If uncertainty in the position of the electron is zero, then its uncertainty in momentum will be:1. h / 2π2. h / 4π3. zero4. infinity

Answer» Correct Answer - Option 4 : infinity

Concept:

Heisenberg’s Uncertainty Principle:

It states that it is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron.
Mathematically, it can be given as

\({\rm{\Delta }}x \times {\rm{\Delta }}{p_x} \ge \frac{h}{{4\pi }}\)

∆x is uncertaintyin position,∆px is uncertaintyin momentum at position x.

Explanation:

Now, it is given thatuncertainty in momentum is zero.

∆px = 0

But, by the uncertainty principle

\({\rm{\Delta }}x \times {\rm{\Delta }}{p_x} \ge \frac{h}{{4\pi }}\)

\(\implies {0} \times {\rm{\Delta }}{x} \ge \frac{h}{{4\pi }}\)

\(\implies {\rm{\Delta }}{x} \ge \frac{h}{{4\pi \times 0 }} \)

Now anything divided by zero is infinity. So,∆x is infinity.

So, infinity is the correct answer.

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