Caesium Stopping Potential

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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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question.rb1 question :caesium_stopping_potential do2   kind :numerical3   describes "The work function of caesium metal is %{work_function}. When light of frequency %{frequency} is " \4             "incident on the metal surface, photoemission of electrons occurs. %{stopping_potential}, in V?"5   given work_function: "2.14 eV"6   given frequency: "6e14 Hz"7   asks stopping_potential: "What is the stopping potential"8 end

solution.rb1 solve :caesium_stopping_potential, using: PhotoelectricEffect do2   objects :caesium3 4   take work_function: caesium.work_function5   take frequency: caesium.frequency6 7   stopping_potential = caesium.stopping_potential8   answer stopping_potential:, in: "V", figures: 29 10   conclusion "The photon's energy minus the work function gives the maximum kinetic energy, which equals the " \11              "energy from the stopping potential."12 end

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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 potential

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 photon's energy minus the work function gives the maximum kinetic energy, which equals the energy from the stopping potential.

The Ruby

1question :caesium_stopping_potential do2  kind :numerical3  describes "The work function of caesium metal is %{work_function}. When light of frequency %{frequency} is " \4            "incident on the metal surface, photoemission of electrons occurs. %{stopping_potential}, in V?"5  given work_function: "2.14 eV"6  given frequency: "6e14 Hz"7  asks stopping_potential: "What is the stopping potential"8end

1solve :caesium_stopping_potential, using: PhotoelectricEffect do2  objects :caesium34  take work_function: caesium.work_function5  take frequency: caesium.frequency67  stopping_potential = caesium.stopping_potential8  answer stopping_potential:, in: "V", figures: 2910  conclusion "The photon's energy minus the work function gives the maximum kinetic energy, which equals the " \11             "energy from the stopping potential."12end

1play_with :caesium_stopping_potential do2  title "Caesium Stopping Potential"3end

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)