Caesium's Stopping Potential

solved PhotoelectricEffect

Q. The work function of caesium metal is 2.14 eV $φ_{0} = 2.14$ eV. When light of frequency 6 × 10¹⁴ Hz $ν = 6 \times 10^{14}$ Hz is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential, in V?

Inputs

work function

φ_{0}

2.14 eV

frequency

ν

6e+14 Hz

Outputs

stopping

V_{0}

How it's solved

Solved with PhotoelectricEffect

Given

$φ_{0} = 2.14$ eV
$ν = 6 \times 10^{14}$ Hz

Find $V_{0}$ .

Working

$φ_{0} = 2.14$ eV $= 3.428658 \times 10^{−19}$ J
einsteins photoelectric equation(caesium) $e \cdot V_{0} = h \cdot ν - φ_{0}$
$1.60217663 \times 10^{−19} \cdot V_{0} = 6.62607015 \times 10^{−34} \cdot 6 \times 10^{14} - 3.428658 \times 10^{−19}$
$V_{0} = 0.3414$ V
$V_{0} \approx 0.34$ Vto 2 significant figures

Conclusion

The stopping potential must take away the fastest electron's energy, which is the photon's energy less the work function.

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_stopping_potential do4  kind :numerical5  describes "The work function of caesium metal is %{work_function}. When light of frequency %{frequency} is " \6            "incident on the metal surface, photoemission of electrons occurs. %{stopping}, in V?"7  given work_function: "2.14 eV"8  given frequency: "6e14 Hz"9  asks stopping: "What is the stopping potential"10end

solution.rb

1require_relative "../questions/caesium_s_stopping_potential"2require_relative "../science/dual_nature_of_radiation_and_matter/photoelectric_effect"34solve :caesium_s_stopping_potential, using: PhotoelectricEffect do5  objects :caesium67  take work_function: caesium.work_function8  take frequency: caesium.frequency910  stopping = caesium.stopping_potential11  answer stopping:, in: "V", figures: 21213  conclusion "The stopping potential must take away the fastest electron's energy, which is the photon's energy " \14             "less the work function."15end

playground.rb

1require_relative "../solutions/caesium_s_stopping_potential"23play_with :caesium_s_stopping_potential do4  title "Caesium's Stopping Potential"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)