Stopping Potential & Maximum KE

PHYPhotoelectric Effect

The stopping potential $V_0$ is the reverse voltage that just halts the most energetic photoelectron, so $eV_0$ equals the maximum kinetic energy $K_{max}$ . Combined with Einstein's photoelectric equation, this lets you find $K_{max}$ , $V_0$ , the electron's maximum speed, or the metal's work function from light of a given wavelength or frequency.

ISC sets these as bread-and-butter numericals, so fluency in switching between eV and joule and between wavelength and frequency is essential.

Einstein's photoelectric equation

K_{max} = h f - \phi_0 = \dfrac{h c}{\lambda} - \phi_0

K_{max}

= maximum KE of photoelectrons,

hf

= incident photon energy,

\phi_0

= work function. All terms in the same unit (J or eV).

Stopping potential

e V_0 = K_{max} \quad\Rightarrow\quad V_0 = \dfrac{h f - \phi_0}{e}

V_0

= stopping potential (V),

e=1.6\times10^{-19}

C. If

\phi_0

and

hf

are in eV, then

V_0

in volts is numerically

K_{max}

in eV.

Maximum speed of photoelectrons

v_{max} = \sqrt{\dfrac{2 K_{max}}{m_e}}

v_{max}

= maximum speed,

m_e=9.1\times10^{-31}

kg.

K_{max}

must be in joule here.

In terms of threshold wavelength

K_{max} = h c\left(\dfrac{1}{\lambda} - \dfrac{1}{\lambda_0}\right)

\lambda

= incident wavelength,

\lambda_0

= threshold wavelength, with

\phi_0=hc/\lambda_0

Stopping potential measures the most energetic electrons only; slower electrons are stopped by smaller voltages, so $V_0$ corresponds to $K_{max}$ , not to an average KE.
Stopping potential is independent of light intensity but increases linearly with frequency; intensity changes only the saturation photocurrent.
Working in eV is fastest: with $hc\approx1240$ eV nm, the photon energy in eV is $1240/\lambda(\text{nm})$ , then $V_0(\text{V}) = K_{max}(\text{eV})$ directly since the charge is one electronic charge.
If $hf<\phi_0$ (i.e. $\lambda>\lambda_0$ ), no emission occurs no matter how intense the light — $K_{max}$ cannot be negative.
To find $\phi_0$ from a measured $V_0$ : $\phi_0 = hf - eV_0 = hc/\lambda - eV_0$ ; keep $eV_0$ and $hc/\lambda$ in the same unit.
Convert $K_{max}$ back to joule before computing $v_{max}$ , and remember it is a maximum speed, not the speed of every electron.
Shorter wavelength (higher frequency) light gives larger $K_{max}$ and larger $V_0$ for the same metal — useful as a quick consistency check between the two-wavelength parts of a question.

Where the marks go

Mixing units inside Einstein's equation — keep $hf$ , $\phi_0$ and $K_{max}$ all in eV or all in joule, never a blend.
Forgetting to convert $K_{max}$ to joule before using $v_{max}=\sqrt{2K_{max}/m_e}$ , which inflates the speed by a huge factor.
Reporting a negative or zero $K_{max}$ as a valid answer when the incident frequency is below threshold — state that no photoemission occurs instead.
Confusing frequency $f$ (Hz) with angular frequency or with wavelength; always check whether the data gives $f$ or $\lambda$ and use the matching form $hf$ or $hc/\lambda$ .

How the board asks it

Numericaleinstein's photoelectric equation and stopping potential
Light of wavelength $4000\,\text{\AA}$ falls on a metal whose work function is $2.0\,\text{eV}$ . Calculate (i) the maximum kinetic energy of the emitted photoelectrons in $\text{eV}$ and (ii) the stopping potential $V_0$ . $(hc = 1240\,\text{eV nm})$
Numericalwork function from a measured stopping potential
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $5000\,\text{\AA}$ is $0.4\,\text{V}$ . Calculate the work function of the metal in $\text{eV}$ . $(hc = 1240\,\text{eV nm})$
Numericalmaximum speed of photoelectrons from $K_{max}$
Radiation of frequency $1.2\times10^{15}\,\text{Hz}$ is incident on a metal of work function $2.5\,\text{eV}$ . Calculate the maximum speed of the ejected photoelectrons. $(h = 6.63\times10^{-34}\,\text{J s},\ m_e = 9.1\times10^{-31}\,\text{kg})$
Diagram / graph $V_0$ varies linearly with frequency; slope gives $h/e$
Sketch a graph showing the variation of stopping potential $V_0$ with the frequency $f$ of incident radiation for a given metal, and explain how the threshold frequency and Planck's constant can be obtained from it.
Give reasons $V_0$ independent of intensity, dependent on frequency
Give reasons: the stopping potential of a photoelectric surface does not change when the intensity of the incident light is increased but does change when its frequency is increased.
Define / state $eV_0 = K_{max}$ for the most energetic electrons
Define stopping potential and state its relation to the maximum kinetic energy $K_{max}$ of the emitted photoelectrons.

Practise this topic

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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.