An α-Particle Turned Back

solved AlphaScattering

Q. In a Geiger-Marsden experiment, a 7.7 MeV $K = 7.7$ MeV α-particle heads straight for a gold nucleus (Z = 79 $Z = 79$ ). What is its distance of closest approach to the nucleus before it comes momentarily to rest and reverses its direction, in metres?

Inputs

energy

K

7.7 MeV

atomic number

Z

Outputs

distance

d

How it's solved

Solved with AlphaScattering

Given

$K = 7.7$ MeV
$Z = 79$

Find $d$ .

Working

$K = 7.7$ MeV $= 1.23367601 \times 10^{−12}$ J
all its energy spent against the nucleus(alpha) $K = \frac{k \cdot 2 \cdot e \cdot Z \cdot e}{d}$
$1.23367601 \times 10^{−12} = \frac{8.98755179 \times 10^{9} \cdot 2 \cdot 1.60217663 \times 10^{−19} \cdot 79 \cdot 1.60217663 \times 10^{−19}}{d}$
$d = 2.955 \times 10^{−14}$ m
$d \approx 3.0 \times 10^{−14}$ mto 2 significant figures

Conclusion

All the α-particle's kinetic energy goes into pushing against the nucleus's charge, so it gets only as close as that energy allows: far smaller than an atom, which is how the nucleus was found.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :an_alpha_particle_turned_back do4  kind :numerical5  describes "In a Geiger-Marsden experiment, a %{energy} α-particle heads straight for a gold nucleus (Z = " \6            "%{atomic_number}). %{distance} before it comes momentarily to rest and reverses its direction, in metres?"7  given energy: "7.7 MeV"8  given atomic_number: "79"9  asks distance: "What is its distance of closest approach to the nucleus"10end

solution.rb

1require_relative "../questions/an_alpha_particle_turned_back"2require_relative "../science/atoms/alpha_scattering"34solve :an_alpha_particle_turned_back, using: AlphaScattering do5  objects :alpha67  take energy: alpha.kinetic_energy8  take atomic_number: alpha.atomic_number910  distance = alpha.closest_approach11  answer distance:, in: "m", figures: 21213  conclusion "All the α-particle's kinetic energy goes into pushing against the nucleus's charge, so it gets " \14             "only as close as that energy allows: far smaller than an atom, which is how the nucleus was found."15end

playground.rb

1require_relative "../solutions/an_alpha_particle_turned_back"23play_with :an_alpha_particle_turned_back do4  title "An α-Particle Turned Back"5end

The laws it drew on

quantity	written	units	within	from
coulomb constant	$k$	N m² C⁻²	—	AlphaScattering
elementary charge	$e$	C	—	AlphaScattering
atomic number	$Z$	—	—	AlphaScattering
kinetic energy	$K$	J	—	AlphaScattering
closest approach	$d$	m	—	AlphaScattering
kinetic energy of ground	$K_{g}$	J	—
atomic number of ground	$Z_{g}$	—	—
closest approach of ground	$d_{g}$	m	—
kinetic energy of alpha	$K$	J	—
atomic number of alpha	$Z$	—	—
closest approach of alpha	$d$	m	—

K_{a} = \frac{k \cdot 2 \cdot e \cdot Z_{a} \cdot e}{d_{a}}

all its energy spent against the nucleus

K_{g} = \frac{k \cdot 2 \cdot e \cdot Z_{g} \cdot e}{d_{g}}

all its energy spent against the nucleus(ground)

K = \frac{k \cdot 2 \cdot e \cdot Z \cdot e}{d}

all its energy spent against the nucleus(alpha)