Logo
Jeeify
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 ms−1\times 10^9 \, \text{ms}^{-1} (nearest integer).

Answer

Correct answer:58

Step-by-step solution

Standard Method

Given: Uncertainty in position of the electron inside the nucleus is Δx=10−15 m\Delta x = 10^{-15} \, \text{m}. Mass of electron is me=9.1×10−31 kgm_e = 9.1 \times 10^{-31} \, \text{kg} and Planck constant is h=6.626×10−34 Jsh = 6.626 \times 10^{-34} \, \text{Js}.

Find: The uncertainty in velocity Δv\Delta v, to be reported as the coefficient of 109 ms−110^9 \, \text{ms}^{-1}.

Using Heisenberg's uncertainty principle,

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

Since

Δp=me⋅Δv\Delta p = m_e \cdot \Delta v

we get

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

Substituting the given values,

10−15⋅9.1×10−31⋅Δv≥6.626×10−344×3.1410^{-15} \cdot 9.1 \times 10^{-31} \cdot \Delta v \geq \frac{6.626 \times 10^{-34}}{4 \times 3.14}

The right-hand side is

6.626×10−3412.56≈5.274×10−35\frac{6.626 \times 10^{-34}}{12.56} \approx 5.274 \times 10^{-35}

Therefore,

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,

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

Expressing this as ×109 ms−1\times 10^9 \, \text{ms}^{-1},

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

Therefore, the required nearest integer is 58.

Direct Substitution

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 Jsh = 6.626 \times 10^{-34} \, \text{Js}.

Find: Δv\Delta v.

Directly use

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

Substitute the values,

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

Simplifying the denominator,

4⋅3.14⋅10−15⋅9.1×10−31=1.143×10−444 \cdot 3.14 \cdot 10^{-15} \cdot 9.1 \times 10^{-31} = 1.143 \times 10^{-44}

Hence,

Δv≥6.626×10−341.143×10−44=5.8×1010 ms−1\Delta v \geq \frac{6.626 \times 10^{-34}}{1.143 \times 10^{-44}} = 5.8 \times 10^{10} \, \text{ms}^{-1}

So,

Δv=58×109 ms−1\Delta v = 58 \times 10^9 \, \text{ms}^{-1}

Therefore, the final answer is 58.

Common mistakes

  • Using Δp=Δv\Delta p = \Delta v is incorrect because momentum uncertainty depends on mass. You must use Δp=meΔv\Delta p = m_e \Delta v for an electron.

  • Forgetting the factor of 4π4\pi in Heisenberg’s uncertainty relation gives a wrong numerical result. Use ΔxΔp≥h4π\Delta x \Delta p \geq \frac{h}{4\pi} exactly as written.

  • Reporting the answer as 5.8×10105.8 \times 10^{10} instead of converting it to the asked form is a presentation error. Rewrite it as 58×10958 \times 10^9 and then give the nearest integer.

Practice more Heisenberg Uncertainty Principle questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Heisenberg Uncertainty Principle

Related questions