Caesium Stopped at 0.60 V

solved PhotoelectricEffect

Q. The work function of caesium is 2.14 eV $φ_{0} = 2.14$ eV. What is the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V $V_{0} = 0.60$ V, in nm?

Inputs

work function

φ_{0}

2.14 eV

stopping

V_{0}

0.6 V

Outputs

wavelength

λ

How it's solved

Solved with PhotoelectricEffect

Given

$φ_{0} = 2.14$ eV
$V_{0} = 0.60$ V

Find $λ$ .

Working

$φ_{0} = 2.14$ eV $= 3.428658 \times 10^{−19}$ J
einsteins photoelectric equation(caesium) $e \cdot V_{0} = h \cdot ν - φ_{0}$
$1.60217663 \times 10^{−19} \cdot 0.6 = 6.62607015 \times 10^{−34} \cdot ν - 3.428658 \times 10^{−19}$
$ν = 6.62529052 \times 10^{14}$ Hz
frequency and wavelength of light(caesium) $c = ν \cdot λ$
$2.99792458 \times 10^{8} = 6.62529052 \times 10^{14} \cdot λ$
$λ = 452.5$ nm
$λ = 4.52497075 \times 10^{−7}$ m $= 452.497075$ nm
$λ \approx 452$ nmto 3 significant figures

Conclusion

The stopping potential measures the fastest electron's energy; add the work function and you have the photon's energy, and so its wavelength.

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_stopped_at_sixty_centivolts do4  kind :numerical5  describes "The work function of caesium is %{work_function}. %{wavelength} if the photocurrent is brought to zero " \6            "by a stopping potential of %{stopping}, in nm?"7  given work_function: "2.14 eV"8  given stopping: "0.60 V"9  asks wavelength: "What is the wavelength of the incident light"10end

solution.rb

1require_relative "../questions/caesium_stopped_at_sixty_centivolts"2require_relative "../science/dual_nature_of_radiation_and_matter/photoelectric_effect"34solve :caesium_stopped_at_sixty_centivolts, using: PhotoelectricEffect do5  objects :caesium67  take work_function: caesium.work_function8  take stopping: caesium.stopping_potential910  wavelength = caesium.wavelength11  answer wavelength:, in: "nm", figures: 31213  conclusion "The stopping potential measures the fastest electron's energy; add the work function and you have " \14             "the photon's energy, and so its wavelength."15end

playground.rb

1require_relative "../solutions/caesium_stopped_at_sixty_centivolts"23play_with :caesium_stopped_at_sixty_centivolts do4  title "Caesium Stopped at 0.60 V"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)