The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) isa)758 fmb)7.58 fmc)75.8 fmd)0.758 fmCorrect answer is option 'B'. Can you explain this answer?

Question

The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) is&#8203;a)758 fmb)7.58 fmc)75.8 fmd)0.758 fmCorrect answer is option 'B'. Can you explain this answer?

Answer

Correct Answer :- b

Explanation : E = 15.0MeV

= 15 * 10⁶ eV

= 15 * 10⁶ * 1.6 * 10^-19 J

= 15 * 1.6 * 10^-13 J

E = (1/4πε_o)*(ze²/r₀²)

r₀ = (1/4πε_o)*(ze²/E)

r₀ = (9*10⁹*79*(1.6*10^-19)2)/(15*1.6*10^-13)

= 75.84 * 10^-16 m

= 7.58 fm

Answer

Answer

Go and see in my dp

Answer

Explanation:
When a proton approaches a gold nucleus, it experiences the electromagnetic force of attraction due to the positive charge of the nucleus. The proton will be repelled by the positive charge of the nucleus until it reaches a certain distance, known as the distance of closest approach.

The distance of closest approach can be calculated using the Rutherford scattering formula, given by:

$$ R = \frac{{kZ_1Z_2e^2}}{{E}} \times \cot(\frac{{\theta}}{2}) $$

Where:
- R is the distance of closest approach in meters.
- k is the Coulomb constant (9.0 x 10^9 Nm^2/C^2)
- Z1 is the atomic number of the incident particle (proton) = 1.
- Z2 is the atomic number of the target nucleus (gold) = 79.
- e is the elementary charge (1.6 x 10^-19 C).
- E is the kinetic energy of the incident particle in Joules.
- θ is the scattering angle.

In this question, the kinetic energy of the proton is given as 15.0 MeV. We need to convert this to Joules using the conversion factor 1 MeV = 1.6 x 10^-13 J.

$$ E = 15.0 \times 1.6 \times 10^{-13} = 2.4 \times 10^{-12} J $$

Substituting the values into the formula:

$$ R = \frac{{9.0 \times 10^9 \times 1 \times 79 \times (1.6 \times 10^{-19})^2}}{{2.4 \times 10^{-12}}} \times \cot(\frac{{\theta}}{2}) $$

The value of the scattering angle θ is not given in the question, so we cannot calculate the exact distance of closest approach. However, we can compare the given options to determine the order of magnitude of the distance.

Comparing the options:
a) 758 fm
b) 7.58 fm
c) 75.8 fm
d) 0.758 fm

Since the atomic radius of a gold nucleus is about 7 fm, option b) 7.58 fm seems plausible as the distance of closest approach. The other options are either too large or too small compared to the atomic radius.

Therefore, the correct answer is option b) 7.58 fm.

Top Courses for Class 12View all

Physics Class 12

ICSE Commerce Class 12 Previous Year Papers

RD Sharma Solutions Mathematics for Class 11 and Class 12

CUET Mock Test: Humanities Subjects 2026

Accountancy Class 12 ICSE

Top Courses for Class 12

Top Courses for Class 12

Related Content

ICSE

CBSE Classes

Competitive Exams