Caesium's Stopping Potential

Q. The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the stopping potential, in V? expects 0.34 V

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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?

The playground

Inputs

work function

φ_{0}

2.14 eV

frequency

ν

6e+14 Hz

Outputs

stopping

V_{0}

The solution

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 Ruby

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

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

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