A Cut-Off Above the Threshold

Q. The threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, what is the cut-off voltage for the photoelectric emission, in V? expects 2.03 V

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

threshold

ν_{0}

3.3e+14 Hz

frequency

ν

8.2e+14 Hz

Outputs

cut off

V_{0}

The solution

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 Ruby

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

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

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

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)