Question

In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV
alpha-particle before it comes momentarily to rest and reverses its direction?

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Answer 1

Answer:

Closest distance of approach of alpha particle towards gold nuclei is

$r = 1.48 \times 10^{-14} m$

Explanation:

In this experiment alpha particles are projected towards the nuclei of gold

so here initial kinetic energy of alpha particle is converted into electrostatic potential energy of alpha particle + gold nuclei

So we will have

$\frac{k(e)(ze)}{r} = KE$

$\frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r} = 7.7 \times 1.6 \times 10^{-13}$

$\frac{1.82 \times 10^{-26}}{r} = 1.23 \times 10^{-12}$

$r = 1.48 \times 10^{-14} m$

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Topic : Electrostatic potential energy

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Answer 2

Explanation:

Answer:

Closest distance of approach of alpha particle towards gold nuclei is

r = 1.48 \times 10^{-14} mr=1.48×10

−14

m

Explanation:

In this experiment alpha particles are projected towards the nuclei of gold

so here initial kinetic energy of alpha particle is converted into electrostatic potential energy of alpha particle + gold nuclei

So we will have

\frac{k(e)(ze)}{r} = KE

r

k(e)(ze)

=KE

\frac{(9\times 10^9)(1.6 \times 10^{-19})^2(79)}{r} = 7.7 \times 1.6 \times 10^{-13}

r

(9×10

9

)(1.6×10

−19

)

2

(79)

=7.7×1.6×10

−13

\frac{1.82 \times 10^{-26}}{r} = 1.23 \times 10^{-12}

r

1.82×10

−26

=1.23×10

−12

r = 1.48 \times 10^{-14} mr=1.48×10

−14

m

#Learn

Topic : Electrostatic potential energy

https://brainly.in/question/12798714