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NVAEasyJEE 2024Heisenberg Uncertainty Principle

JEE Chemistry 2024 Question with Solution

Based on Heisenberg’s uncertainty principle, the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter 10−15 m10^{-15} \, \text{m} is ........ × 109 m s−110^9 \, \text{m s}^{-1} (nearest integer).

Answer

Correct answer:58

Step-by-step solution

Standard Method

Given: Uncertainty in position, Δx=10−15 m\Delta x = 10^{-15} \, \text{m}; mass of electron, me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \, \text{kg}; Planck constant, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \, \text{J s}.

Find: The uncertainty in velocity Δv\Delta v in the form _____×109 m s−1\_\_\_\_\_ \times 10^9 \, \text{m s}^{-1}.

Using Heisenberg's uncertainty principle:

Δx⋅Δp≥h4π\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

Since

Δp=meΔv\Delta p = m_e \Delta v

we get

Δx⋅me⋅Δv≥h4π\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}

Therefore,

Δv≥h4πΔxme\Delta v \geq \frac{h}{4\pi \Delta x m_e}

Substituting the given values:

Δv≥6.626×10−344×3.14×10−15×9.1×10−31\Delta v \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 10^{-15} \times 9.1 \times 10^{-31}}

Simplifying,

Δv≥5.8×1010 m s−1\Delta v \geq 5.8 \times 10^{10} \, \text{m s}^{-1}

Now write it as

5.8×1010=58×1095.8 \times 10^{10} = 58 \times 10^9

Therefore, the uncertainty in velocity is 58×109 m s−158 \times 10^9 \, \text{m s}^{-1}, so the final answer is 58.

Detailed Calculation

Given: Δx=10−15 m\Delta x = 10^{-15} \, \text{m}, me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \, \text{kg}, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \, \text{J s}.

Find: The nearest integer in Δv=(_____)×109 m s−1\Delta v = (\_\_\_\_\_) \times 10^9 \, \text{m s}^{-1}.

Start from

Δx⋅me⋅Δv≥h4π\Delta x \cdot m_e \cdot \Delta v \geq \frac{h}{4\pi}

First evaluate the right-hand side:

6.626×10−344×3.14≈5.274×10−35\frac{6.626 \times 10^{-34}}{4 \times 3.14} \approx 5.274 \times 10^{-35}

Now,

10−15⋅9.1×10−31⋅Δv≥5.274×10−3510^{-15} \cdot 9.1 \times 10^{-31} \cdot \Delta v \geq 5.274 \times 10^{-35}

So,

9.1×10−46⋅Δv≥5.274×10−359.1 \times 10^{-46} \cdot \Delta v \geq 5.274 \times 10^{-35}

Hence,

Δv≥5.274×10−359.1×10−46=5.796×1010 m s−1\Delta v \geq \frac{5.274 \times 10^{-35}}{9.1 \times 10^{-46}} = 5.796 \times 10^{10} \, \text{m s}^{-1}

Rounding to the required form,

5.796×1010≈58×109 m s−15.796 \times 10^{10} \approx 58 \times 10^9 \, \text{m s}^{-1}

Therefore, the final answer is 58.

Common mistakes

  • Using diameter incorrectly by taking Δx\Delta x larger or smaller than the given 10−15 m10^{-15} \, \text{m} without following the question statement. Here the positional uncertainty is taken from the stated nuclear size, so substitute the given value directly.

  • Forgetting that Δp=meΔv\Delta p = m_e \Delta v for a non-relativistic substitution step. If momentum uncertainty is not converted to velocity uncertainty using electron mass, the final quantity will not match what is asked.

  • Making an exponent error while dividing powers of 1010. Since 10−34/10−44=101010^{-34} / 10^{-44} = 10^{10}, a sign mistake here can change the answer by many orders of magnitude.

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