Threshold Frequency & Wavelength | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Threshold frequency and wavelength mark the boundary

Imagine increasing the radiation frequency from a low value. At first, no electrons escape. At one boundary frequency, emission just begins.

Definition

1 mark

What is meant by the threshold frequency?

Model answer: The threshold frequency is the minimum frequency of electromagnetic radiation that causes electrons to be emitted from the surface of a metal.

Definition

1 mark

What is meant by the threshold wavelength?

Model answer: The threshold wavelength is the maximum wavelength of electromagnetic radiation that causes electrons to be emitted from the surface of a metal.

At the threshold, the fastest emitted electron has zero kinetic energy. The entire photon energy is used to escape.

\Phi=hf_0

Symbols

$\Phi$ = work function of the metal (J)
$f_0$ = threshold frequency for the metal (Hz)
$h$ = Planck constant (J s)

The threshold frequency has a corresponding threshold wavelength. Using the wave equation gives the wavelength form below.

\lambda_0=\dfrac{hc}{\Phi}

Symbols

$\lambda_0$ = threshold wavelength for the metal (m)
$h$ = Planck constant (J s)
$c$ = wave speed in a vacuum (m s⁻¹)
$\Phi$ = work function of the metal (J)

Worked example

Find the largest threshold wavelength

Potassium has a work function of 2.26 eV. Calculate its threshold wavelength (9702/42/F/M/24 Q7(c)).

Convert the work function to joules.
$\Phi=2.26\ \mathrm{eV}$
$\Phi=3.62\times10^{-19}\ \mathrm{J}$
At the threshold, set the photon energy equal to the work function.
$hc=1.99\times10^{-25}\ \mathrm{J\,m}$
$\lambda_0=\dfrac{hc}{\Phi}$
$\lambda_0=5.50\times10^{-7}\ \mathrm{m}$

Answer

\lambda_0=5.50\times10^{-7}\ \mathrm{m}

Your turnrelationship check

Metal B has a larger work function than metal A. Compare the threshold frequency and threshold wavelength of B with those of A. [2]

Show worked answer

B has a higher threshold frequency because more photon energy is needed. Its threshold wavelength is therefore shorter.

A larger work function needs a higher threshold frequency and gives a shorter threshold wavelength.

Threshold frequency and wavelength mark the boundary

Imagine increasing the radiation frequency from a low value. At first, no electrons escape. At one boundary frequency, emission just begins.

Definition

1 mark

What is meant by the threshold frequency?

Model answer: The threshold frequency is the minimum frequency of electromagnetic radiation that causes electrons to be emitted from the surface of a metal.

Definition

1 mark

What is meant by the threshold wavelength?

Model answer: The threshold wavelength is the maximum wavelength of electromagnetic radiation that causes electrons to be emitted from the surface of a metal.

At the threshold, the fastest emitted electron has zero kinetic energy. The entire photon energy is used to escape.

\Phi=hf_0

Symbols

$\Phi$ = work function of the metal (J)
$f_0$ = threshold frequency for the metal (Hz)
$h$ = Planck constant (J s)

The threshold frequency has a corresponding threshold wavelength. Using the wave equation gives the wavelength form below.

\lambda_0=\dfrac{hc}{\Phi}

Symbols

$\lambda_0$ = threshold wavelength for the metal (m)
$h$ = Planck constant (J s)
$c$ = wave speed in a vacuum (m s⁻¹)
$\Phi$ = work function of the metal (J)

Worked example

Find the largest threshold wavelength

Potassium has a work function of 2.26 eV. Calculate its threshold wavelength (9702/42/F/M/24 Q7(c)).

Convert the work function to joules.
$\Phi=2.26\ \mathrm{eV}$
$\Phi=3.62\times10^{-19}\ \mathrm{J}$
At the threshold, set the photon energy equal to the work function.
$hc=1.99\times10^{-25}\ \mathrm{J\,m}$
$\lambda_0=\dfrac{hc}{\Phi}$
$\lambda_0=5.50\times10^{-7}\ \mathrm{m}$

Answer

\lambda_0=5.50\times10^{-7}\ \mathrm{m}

Your turnrelationship check

Metal B has a larger work function than metal A. Compare the threshold frequency and threshold wavelength of B with those of A. [2]

Show worked answer

B has a higher threshold frequency because more photon energy is needed. Its threshold wavelength is therefore shorter.

A larger work function needs a higher threshold frequency and gives a shorter threshold wavelength.