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Calculate the distance of closest approach between an alpha particle of energy 5.5 MeV (5.5 × 10^6 × 1.6 × 10^-19) if the charge on the gold nucleus is 79e = 79 × 1.6 × 10^-19 C and the charge on the alpha particle is 2e = 2 × 1.6 × 10^-19 C?
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Calculate the distance of closest approach between an alpha particle o...
Given:
Energy of the alpha particle (E) = 5.5 MeV = 5.5 × 10^6 × 1.6 × 10^-19 J
Charge on the gold nucleus (Q) = 79e = 79 × 1.6 × 10^-19 C
Charge on the alpha particle (q) = 2e = 2 × 1.6 × 10^-19 C

To find:
Distance of closest approach between the alpha particle and the gold nucleus.

Formula:
The distance of closest approach (r) can be calculated using the Coulomb's law:

E = (Qq)/(4πε₀r)

Where:
E = Energy of the alpha particle
Q = Charge on the gold nucleus
q = Charge on the alpha particle
ε₀ = Permittivity of free space
r = Distance of closest approach

Solution:
Step 1: Convert the energy of the alpha particle from MeV to joules:
5.5 MeV = 5.5 × 10^6 × 1.6 × 10^-19 J

Step 2: Substitute the given values into the formula:
5.5 × 10^6 × 1.6 × 10^-19 J = (79 × 1.6 × 10^-19 C × 2 × 1.6 × 10^-19 C)/(4πε₀r)

Step 3: Simplify the equation:
5.5 × 10^6 × 1.6 × 10^-19 J = (79 × 1.6 × 10^-19 C × 2 × 1.6 × 10^-19 C)/(4πε₀r)

Step 4: Cancel out common terms:
5.5 × 10^6 = 79 × 2/(4πε₀r)

Step 5: Rearrange the equation to solve for r:
r = (79 × 2)/(5.5 × 10^6 × 4πε₀)

Step 6: Substitute the values of ε₀ and π:
ε₀ = 8.854 × 10^-12 C^2/(N·m^2)
π ≈ 3.14159

Step 7: Calculate the value of r:
r = (79 × 2)/(5.5 × 10^6 × 4 × 3.14159 × 8.854 × 10^-12) m

Step 8: Simplify the expression:
r ≈ 1.44 × 10^-14 m

Answer:
The distance of closest approach between the alpha particle and the gold nucleus is approximately 1.44 × 10^-14 meters.
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Calculate the distance of closest approach between an alpha particle of energy 5.5 MeV (5.5 × 10^6 × 1.6 × 10^-19) if the charge on the gold nucleus is 79e = 79 × 1.6 × 10^-19 C and the charge on the alpha particle is 2e = 2 × 1.6 × 10^-19 C?
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Calculate the distance of closest approach between an alpha particle of energy 5.5 MeV (5.5 × 10^6 × 1.6 × 10^-19) if the charge on the gold nucleus is 79e = 79 × 1.6 × 10^-19 C and the charge on the alpha particle is 2e = 2 × 1.6 × 10^-19 C? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Calculate the distance of closest approach between an alpha particle of energy 5.5 MeV (5.5 × 10^6 × 1.6 × 10^-19) if the charge on the gold nucleus is 79e = 79 × 1.6 × 10^-19 C and the charge on the alpha particle is 2e = 2 × 1.6 × 10^-19 C? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the distance of closest approach between an alpha particle of energy 5.5 MeV (5.5 × 10^6 × 1.6 × 10^-19) if the charge on the gold nucleus is 79e = 79 × 1.6 × 10^-19 C and the charge on the alpha particle is 2e = 2 × 1.6 × 10^-19 C?.
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