X-Rays From Thirty Kilovolt Electrons

solved Photon

Q. X-rays are produced by $E = e \cdot V$ 30 kV $V = 30$ kV electrons. What is the maximum frequency of the X-rays, in Hz?

Inputs

voltage

V

30 kV

Outputs

frequency

ν

How it's solved

Solved with Photon

Given

$V = 30$ kV
$E = e \cdot V$ from “produced by”

Find $ν$ .

Working

$V = 30$ kV $= 30000$ V
produced by $E = e \cdot V$
$E = 1.60217663 \times 10^{−19} \cdot 30000$
$E = 4.8065299 \times 10^{−15}$ J
energy of a photon(x rays) $E = h \cdot ν$
$4.8065299 \times 10^{−15} = 6.62607015 \times 10^{−34} \cdot ν$
$ν = 7.254 \times 10^{18}$ Hz
$ν \approx 7.25 \times 10^{18}$ Hzto 3 significant figures

Conclusion

The most energetic X-ray photon takes all the energy an electron gains across the accelerating voltage, which sets the highest frequency.

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 :x_rays_from_thirty_kilovolt_electrons do4  kind :numerical5  describes "X-rays are produced by %{voltage} electrons. %{frequency}, in Hz?"6  given voltage: "30 kV"7  asks frequency: "What is the maximum frequency of the X-rays"8end

solution.rb

1require_relative "../questions/x_rays_from_thirty_kilovolt_electrons"2require_relative "../science/dual_nature_of_radiation_and_matter/photon"34solve :x_rays_from_thirty_kilovolt_electrons, using: Photon do5  objects :x_rays67  take voltage: x_rays.accelerating_potential89  deduce_from "produced by", x_rays.photon_energy => e * x_rays.accelerating_potential1011  frequency = x_rays.frequency12  answer frequency:, in: "Hz", figures: 31314  conclusion "The most energetic X-ray photon takes all the energy an electron gains across the accelerating " \15             "voltage, which sets the highest frequency."16end

playground.rb

1require_relative "../solutions/x_rays_from_thirty_kilovolt_electrons"23play_with :x_rays_from_thirty_kilovolt_electrons do4  title "X-Rays From Thirty Kilovolt Electrons"5end

The laws it drew on

quantity	written	units	within	from
plancks constant	$h$	J s	—	Photon
speed of light	$c$	m/s	—	Photon
elementary charge	$e$	C	—	Photon
frequency	$ν$	Hz	0 … ∞	Photon
wavelength	$λ$	m	0 … ∞	Photon
photon energy	$E$	J	—	Photon
photon momentum	$p$	kg m/s	—	Photon
emitted power	$P$	W	—	Photon
photons a second	$N$	s⁻¹	—	Photon
accelerating potential	$V$	V	—	Photon
photon energy of ground	$E_{g}$	J	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
photon energy of x rays	$E$	J	—
frequency of x rays	$ν$	Hz	0 … ∞
wavelength of ground	$λ_{g}$	m	0 … ∞
wavelength of x rays	$λ$	m	0 … ∞
photon momentum of ground	$p_{g}$	kg m/s	—
photon momentum of x rays	$p$	kg m/s	—
emitted power of ground	$P_{g}$	W	—
photons a second of ground	$N_{g}$	s⁻¹	—
emitted power of x rays	$P$	W	—
photons a second of x rays	$N$	s⁻¹	—
accelerating potential of x rays	$V$	V	—

E_{l} = h \cdot ν_{l}

energy of a photon

c = ν_{l} \cdot λ_{l}

frequency and wavelength of light

p_{l} = \frac{h}{λ_{l}}

momentum of a photon

P_{l} = N_{l} \cdot E_{l}

photons carry the power

E_{g} = h \cdot ν_{g}

energy of a photon(ground)

E = h \cdot ν

energy of a photon(x rays)

c = ν_{g} \cdot λ_{g}

frequency and wavelength of light(ground)

c = ν \cdot λ

frequency and wavelength of light(x rays)

p_{g} = \frac{h}{λ_{g}}

momentum of a photon(ground)

p = \frac{h}{λ}

momentum of a photon(x rays)

P_{g} = N_{g} \cdot E_{g}

photons carry the power(ground)

P = N \cdot E

photons carry the power(x rays)

E = e \cdot V

produced by