Caesium's Threshold

solved PhotoelectricEffect

Q. The work function of caesium is 2.14 eV $φ_{0} = 2.14$ eV. What is the threshold frequency for caesium, in hertz?

Inputs

work function

φ_{0}

2.14 eV

Outputs

threshold

ν_{0}

How it's solved

Solved with PhotoelectricEffect

Given

$φ_{0} = 2.14$ eV

Find $ν_{0}$ .

Working

$φ_{0} = 2.14$ eV $= 3.428658 \times 10^{−19}$ J
threshold of emission(caesium) $h \cdot ν_{0} = φ_{0}$
$6.62607015 \times 10^{−34} \cdot ν_{0} = 3.428658 \times 10^{−19}$
$ν_{0} = 5.174 \times 10^{14}$ Hz
$ν_{0} \approx 5.17 \times 10^{14}$ Hzto 3 significant figures

Conclusion

A photon frees an electron only if it brings at least the work function; that sets the lowest frequency that can do it.

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 :caesium_s_threshold do4  kind :numerical5  describes "The work function of caesium is %{work_function}. %{threshold}, in hertz?"6  given work_function: "2.14 eV"7  asks threshold: "What is the threshold frequency for caesium"8end

solution.rb

1require_relative "../questions/caesium_s_threshold"2require_relative "../science/dual_nature_of_radiation_and_matter/photoelectric_effect"34solve :caesium_s_threshold, using: PhotoelectricEffect do5  objects :caesium67  take work_function: caesium.work_function89  threshold = caesium.threshold_frequency10  answer threshold:, in: "Hz", figures: 31112  conclusion "A photon frees an electron only if it brings at least the work function; that sets the lowest " \13             "frequency that can do it."14end

playground.rb

1require_relative "../solutions/caesium_s_threshold"23play_with :caesium_s_threshold do4  title "Caesium's Threshold"5end

The laws it drew on

quantity	written	units	within	from
plancks constant	$h$	J s	—	PhotoelectricEffect
elementary charge	$e$	C	—	PhotoelectricEffect
speed of light	$c$	m/s	—	PhotoelectricEffect
frequency	$ν$	Hz	0 … ∞	PhotoelectricEffect
wavelength	$λ$	m	0 … ∞	PhotoelectricEffect
work function	$φ_{0}$	J	—	PhotoelectricEffect
threshold frequency	$ν_{0}$	Hz	—	PhotoelectricEffect
stopping potential	$V_{0}$	V	—	PhotoelectricEffect
maximum kinetic energy	$K_{\max}$	J	—	PhotoelectricEffect
frequency of ground	$ν_{g}$	Hz	0 … ∞
wavelength of ground	$λ_{g}$	m	0 … ∞
frequency of caesium	$ν$	Hz	0 … ∞
wavelength of caesium	$λ$	m	0 … ∞
threshold frequency of ground	$ν_{0,g}$	Hz	—
work function of ground	$φ_{0,g}$	J	—
threshold frequency of caesium	$ν_{0}$	Hz	—
work function of caesium	$φ_{0}$	J	—
stopping potential of ground	$V_{0,g}$	V	—
stopping potential of caesium	$V_{0}$	V	—
maximum kinetic energy of ground	$K_{max,g}$	J	—
maximum kinetic energy of caesium	$K_{\max}$	J	—

c = ν_{l} \cdot λ_{l}

frequency and wavelength of light

h \cdot ν_{0,m} = φ_{0,m}

threshold of emission

e \cdot V_{0,m} = h \cdot ν_{m} - φ_{0,m}

einsteins photoelectric equation

e \cdot V_{0,m} = K_{max,m}

the stopping potential stops the fastest

c = ν_{g} \cdot λ_{g}

frequency and wavelength of light(ground)

c = ν \cdot λ

frequency and wavelength of light(caesium)

h \cdot ν_{0,g} = φ_{0,g}

threshold of emission(ground)

h \cdot ν_{0} = φ_{0}

threshold of emission(caesium)

e \cdot V_{0,g} = h \cdot ν_{g} - φ_{0,g}

einsteins photoelectric equation(ground)

e \cdot V_{0} = h \cdot ν - φ_{0}

einsteins photoelectric equation(caesium)

e \cdot V_{0,g} = K_{max,g}

the stopping potential stops the fastest(ground)

e \cdot V_{0} = K_{\max}

the stopping potential stops the fastest(caesium)