A Cut-Off of One and a Half Volts

solved PhotoelectricEffect

Q. The photoelectric cut-off voltage in a certain experiment is 1.5 V $V_{0} = 1.5$ V. What is the maximum kinetic energy of photoelectrons emitted, in J?

Inputs

cut off

V_{0}

1.5 V

Outputs

energy

K_{\max}

How it's solved

Solved with PhotoelectricEffect

Given

$V_{0} = 1.5$ V

Find $K_{\max}$ .

Working

the stopping potential stops the fastest(emitter) $e \cdot V_{0} = K_{\max}$
$1.60217663 \times 10^{−19} \cdot 1.5 = K_{\max}$
$K_{\max} = 2.403 \times 10^{−19}$ J
$K_{\max} \approx 2.4 \times 10^{−19}$ Jto 2 significant figures

Conclusion

The cut-off voltage just stops the fastest electrons, so their kinetic energy is the electron's charge times that voltage.

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 :a_cut_off_of_one_and_a_half_volts do4  kind :numerical5  describes "The photoelectric cut-off voltage in a certain experiment is %{cut_off}. %{energy}, in J?"6  given cut_off: "1.5 V"7  asks energy: "What is the maximum kinetic energy of photoelectrons emitted"8end

solution.rb

1require_relative "../questions/a_cut_off_of_one_and_a_half_volts"2require_relative "../science/dual_nature_of_radiation_and_matter/photoelectric_effect"34solve :a_cut_off_of_one_and_a_half_volts, using: PhotoelectricEffect do5  objects :emitter67  take cut_off: emitter.stopping_potential89  energy = emitter.maximum_kinetic_energy10  answer energy:, in: "J", figures: 21112  conclusion "The cut-off voltage just stops the fastest electrons, so their kinetic energy is the electron's " \13             "charge times that voltage."14end

playground.rb

1require_relative "../solutions/a_cut_off_of_one_and_a_half_volts"23play_with :a_cut_off_of_one_and_a_half_volts do4  title "A Cut-Off of One and a Half Volts"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 emitter	$ν$	Hz	0 … ∞
wavelength of emitter	$λ$	m	0 … ∞
threshold frequency of ground	$ν_{0,g}$	Hz	—
work function of ground	$φ_{0,g}$	J	—
threshold frequency of emitter	$ν_{0}$	Hz	—
work function of emitter	$φ_{0}$	J	—
stopping potential of ground	$V_{0,g}$	V	—
stopping potential of emitter	$V_{0}$	V	—
maximum kinetic energy of ground	$K_{max,g}$	J	—
maximum kinetic energy of emitter	$K_{\max}$	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(emitter)

h \cdot ν_{0,g} = φ_{0,g}

threshold of emission(ground)

h \cdot ν_{0} = φ_{0}

threshold of emission(emitter)

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(emitter)

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(emitter)