An α-Particle Turned Back

Q. In a Geiger-Marsden experiment, a 7.7 MeV α-particle heads straight for a gold nucleus (Z = 79). What is its distance of closest approach to the nucleus before it comes momentarily to rest and reverses its direction, in metres? expects 3.0 × 10⁻¹⁴ m

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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?

The playground

Inputs

energy

K

7.7 MeV

atomic number

Z

Outputs

distance

d

The solution

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 Ruby

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

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

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

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)