A Cut-Off Above the Threshold

solved PhotoelectricEffect

Q. The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz $ν_{0} = 3.3 \times 10^{14}$ Hz. If light of frequency 8.2 × 10¹⁴ Hz $ν = 8.2 \times 10^{14}$ Hz is incident on the metal, what is the cut-off voltage for the photoelectric emission, in V?

Inputs

threshold

ν_{0}

3.3e+14 Hz

frequency

ν

8.2e+14 Hz

Outputs

cut off

V_{0}

How it's solved

Solved with PhotoelectricEffect

Given

$ν_{0} = 3.3 \times 10^{14}$ Hz
$ν = 8.2 \times 10^{14}$ Hz

Find $V_{0}$ .

Working

threshold of emission(metal) $h \cdot ν_{0} = φ_{0}$
$6.62607015 \times 10^{−34} \cdot 3.3 \times 10^{14} = φ_{0}$
$φ_{0} = 2.18660315 \times 10^{−19}$ J
einsteins photoelectric equation(metal) $e \cdot V_{0} = h \cdot ν - φ_{0}$
$1.60217663 \times 10^{−19} \cdot V_{0} = 6.62607015 \times 10^{−34} \cdot 8.2 \times 10^{14} - 2.18660315 \times 10^{−19}$
$V_{0} = 2.026$ V
$V_{0} \approx 2.03$ Vto 3 significant figures

Conclusion

Only the frequency beyond the threshold goes into the electrons' motion, and the cut-off voltage must undo that energy.

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_above_the_threshold do4  kind :numerical5  describes "The threshold frequency for a certain metal is %{threshold}. If light of frequency %{frequency} is " \6            "incident on the metal, %{cut_off}, in V?"7  given threshold: "3.3e14 Hz"8  given frequency: "8.2e14 Hz"9  asks cut_off: "what is the cut-off voltage for the photoelectric emission"10end

solution.rb

1require_relative "../questions/a_cut_off_above_the_threshold"2require_relative "../science/dual_nature_of_radiation_and_matter/photoelectric_effect"34solve :a_cut_off_above_the_threshold, using: PhotoelectricEffect do5  objects :metal67  take threshold: metal.threshold_frequency8  take frequency: metal.frequency910  cut_off = metal.stopping_potential11  answer cut_off:, in: "V", figures: 31213  conclusion "Only the frequency beyond the threshold goes into the electrons' motion, and the cut-off voltage " \14             "must undo that energy."15end

playground.rb

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

h \cdot ν_{0,g} = φ_{0,g}

threshold of emission(ground)

h \cdot ν_{0} = φ_{0}

threshold of emission(metal)

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(metal)

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(metal)