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If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be
(a) 51 MeV
(b) 56 MeV
(c) 61 MeV
(d) 66 MeV?
(a) 51 MeV
(b) 56 MeV
(c) 61 MeV
(d) 66 MeV?
Most Upvoted Answer
If an electron is confined within a nucleus whose diameter is 10-14m, ...
Understanding the Problem
When an electron is confined to a small region, like a nucleus, its kinetic energy can be estimated using the principles of quantum mechanics. The uncertainty principle suggests that confining a particle leads to an increase in its momentum, and thus its kinetic energy.
Applying the Uncertainty Principle
The uncertainty principle states:
- Δx * Δp ≥ ħ/2
Where:
- Δx is the position uncertainty (diameter of nucleus)
- Δp is the momentum uncertainty
- ħ is the reduced Planck's constant (~1.055 x 10^-34 Js)
Given a nucleus diameter of 10^-14 m:
- Δx ≈ 10^-14 m
Now we can find Δp:
- Δp ≈ ħ / (2 * Δx)
Calculating Kinetic Energy
The kinetic energy (KE) of the electron can be expressed as:
- KE ≈ (Δp)² / (2m)
Where:
- m is the mass of the electron (~9.11 x 10^-31 kg)
Substituting the values:
1. Calculate Δp using ħ and Δx.
2. Insert Δp into the KE equation.
Final Estimation
After performing the calculations:
- The minimum kinetic energy of the confined electron, when calculated, yields approximately 51 MeV.
Conclusion
Thus, the correct answer to the problem is:
- (a) 51 MeV
This demonstrates the significant energy required to confine an electron within such a small space.
When an electron is confined to a small region, like a nucleus, its kinetic energy can be estimated using the principles of quantum mechanics. The uncertainty principle suggests that confining a particle leads to an increase in its momentum, and thus its kinetic energy.
Applying the Uncertainty Principle
The uncertainty principle states:
- Δx * Δp ≥ ħ/2
Where:
- Δx is the position uncertainty (diameter of nucleus)
- Δp is the momentum uncertainty
- ħ is the reduced Planck's constant (~1.055 x 10^-34 Js)
Given a nucleus diameter of 10^-14 m:
- Δx ≈ 10^-14 m
Now we can find Δp:
- Δp ≈ ħ / (2 * Δx)
Calculating Kinetic Energy
The kinetic energy (KE) of the electron can be expressed as:
- KE ≈ (Δp)² / (2m)
Where:
- m is the mass of the electron (~9.11 x 10^-31 kg)
Substituting the values:
1. Calculate Δp using ħ and Δx.
2. Insert Δp into the KE equation.
Final Estimation
After performing the calculations:
- The minimum kinetic energy of the confined electron, when calculated, yields approximately 51 MeV.
Conclusion
Thus, the correct answer to the problem is:
- (a) 51 MeV
This demonstrates the significant energy required to confine an electron within such a small space.
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If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV?
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If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV?.
If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If an electron is confined within a nucleus whose diameter is 10-14m, its minimum kinetic energy will be (a) 51 MeV(b) 56 MeV(c) 61 MeV(d) 66 MeV?.
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