Caesium's Threshold

Q. The work function of caesium is 2.14 eV. What is the threshold frequency for caesium, in hertz? expects 5.17 × 10¹⁴ Hz

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Q. The work function of caesium is 2.14 eV $φ_{0} = 2.14$ eV. What is the threshold frequency for caesium, in hertz?

The playground

Inputs

work function

φ_{0}

2.14 eV

Outputs

threshold

ν_{0}

The solution

Solved with PhotoelectricEffect

Given

$φ_{0} = 2.14$ eV

Find $ν_{0}$ .

Working

$φ_{0} = 2.14$ eV $= 3.428658 \times 10^{−19}$ J
threshold of emission(caesium) $h \cdot ν_{0} = φ_{0}$
$6.62607015 \times 10^{−34} \cdot ν_{0} = 3.428658 \times 10^{−19}$
$ν_{0} = 5.174 \times 10^{14}$ Hz
$ν_{0} \approx 5.17 \times 10^{14}$ Hzto 3 significant figures

Conclusion

A photon frees an electron only if it brings at least the work function; that sets the lowest frequency that can do it.

The Ruby

1require_relative "../physics"23question :caesium_s_threshold do4  kind :numerical5  describes "The work function of caesium is %{work_function}. %{threshold}, in hertz?"6  given work_function: "2.14 eV"7  asks threshold: "What is the threshold frequency for caesium"8end

1require_relative "../questions/caesium_s_threshold"2require_relative "../science/dual_nature_of_radiation_and_matter/photoelectric_effect"34solve :caesium_s_threshold, using: PhotoelectricEffect do5  objects :caesium67  take work_function: caesium.work_function89  threshold = caesium.threshold_frequency10  answer threshold:, in: "Hz", figures: 31112  conclusion "A photon frees an electron only if it brings at least the work function; that sets the lowest " \13             "frequency that can do it."14end

1require_relative "../solutions/caesium_s_threshold"23play_with :caesium_s_threshold do4  title "Caesium's 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 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)