A Cut-Off of One and a Half Volts

Q. The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted, in J? expects 2.4 × 10⁻¹⁹ J

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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?

The playground

Inputs

cut off

V_{0}

1.5 V

Outputs

energy

K_{\max}

The solution

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 Ruby

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

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

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

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)