Dual Nature of Radiation and Matter · Study notes

PHYPhotoelectric Effect

Work Function & Threshold

The work function $\phi_0$ is the minimum energy needed to free an electron from a metal surface, and it fixes the threshold frequency $f_0$ and threshold wavelength $\lambda_0$ below/above which no photoemission occurs. These quantities link directly through $\phi_0=hf_0=hc/\lambda_0$ , and ISC questions ask you to compute one from the others and to decide whether a given light will eject electrons.

The key judgement is comparing incident frequency with $f_0$ (or incident wavelength with $\lambda_0$ ).

Work function and threshold

\phi_0 = h f_0 = \dfrac{h c}{\lambda_0}

\phi_0

= work function (J or eV),

f_0

= threshold frequency (Hz),

\lambda_0

= threshold wavelength (m).

Threshold frequency

f_0 = \dfrac{\phi_0}{h}

f_0

= minimum frequency for emission;

h=6.63\times10^{-34}

J s. Use

\phi_0

in joule.

Threshold wavelength

\lambda_0 = \dfrac{h c}{\phi_0} = \dfrac{c}{f_0}

\lambda_0

= longest wavelength that can eject electrons. Shortcut:

\lambda_0(\text{nm}) \approx 1240/\phi_0(\text{eV})

Emission condition

f \ge f_0 \quad \Longleftrightarrow \quad \lambda \le \lambda_0

f,\lambda

= incident frequency and wavelength. Photoemission occurs only when this is satisfied.

Work function is a property of the metal alone; it is independent of the incident light and is usually quoted in eV (convert with $1\,\text{eV}=1.6\times10^{-19}$ J).
Emission needs frequency at or above $f_0$ , equivalently wavelength at or below $\lambda_0$ — note the inequality flips because $\lambda$ and $f$ are inversely related.
The shortcut $\lambda_0(\text{nm})=1240/\phi_0(\text{eV})$ lets you check answers fast: a 2.3 eV metal has $\lambda_0\approx 539$ nm.
A larger work function means a higher threshold frequency and a shorter threshold wavelength — harder to eject electrons.
To decide if a given light causes emission, compare its frequency with $f_0$ (or its wavelength with $\lambda_0$ ); intensity is irrelevant to whether emission happens at all.
Threshold corresponds to zero kinetic energy of the ejected electron: at exactly $f=f_0$ , $K_{max}=0$ and the stopping potential is zero.
Keep $hc\approx1.986\times10^{-25}$ J m (or $1240$ eV nm) handy; choosing the right value of $hc$ avoids unit slips between SI and eV-nm working.

Where the marks go

Flipping the emission condition: students sometimes write $\lambda \ge \lambda_0$ for emission; the correct rule is $\lambda \le \lambda_0$ (and $f \ge f_0$ ).
Leaving $\phi_0$ in eV when dividing by $h$ to get $f_0$ — convert to joule first or you will be off by $1.6\times10^{-19}$ .
Confusing threshold frequency $f_0$ with the incident frequency; $f_0$ is a fixed metal property, the incident $f$ is what you compare against it.
Mixing the two $hc$ values — using $1240$ eV nm with wavelengths in metres, or $1.986\times10^{-25}$ J m with energies in eV.

How the board asks it

Numericalthe relation $\phi_0=hf_0=hc/\lambda_0$
The work function of a metal is $2.3\,\text{eV}$ . Calculate the threshold frequency $f_0$ and the threshold wavelength $\lambda_0$ for this metal. $(h=6.63\times10^{-34}\,\text{J s},\ c=3\times10^8\,\text{m s}^{-1},\ 1\,\text{eV}=1.6\times10^{-19}\,\text{J})$
Give reasonsemission condition $f\ge f_0$
Light of wavelength $600\,\text{nm}$ is incident on a metal of work function $2.5\,\text{eV}$ . Will photoelectrons be emitted? Justify your answer with a calculation.
Define / statework function as a property of the metal
Define the term work function of a metal surface and state the unit in which it is commonly expressed.
Give reasonsintensity is irrelevant to whether emission happens
A metal surface emits no photoelectrons when illuminated by intense red light, yet a feeble beam of violet light causes emission. Account for this observation in terms of the threshold frequency.
Assertion–Reasonlarger work function gives a shorter threshold wavelength
Assertion: A metal with a larger work function has a shorter threshold wavelength. Reason: The threshold wavelength is given by $\lambda_0=hc/\phi_0$ . Choose the correct option regarding these two statements.
Numericalthreshold corresponds to zero kinetic energy
The threshold wavelength for a metal is $540\,\text{nm}$ . Calculate its work function in $\text{eV}$ , and state the maximum kinetic energy of the electrons ejected when light of exactly this wavelength falls on it.

Practise this topic

NextStopping Potential & Maximum KE

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.