Q.The work function of caesium is $2.14\ \text{eV}$. Find

Einstein's photoelectric equation: $h\nu=\phi+eV_0$.

(a) Threshold frequency: at threshold $K_{\max}=0$, so $\nu_0=\phi/h = \dfrac{2.14\ \text{eV}}{4.1357\times10^{-15}\ \text{eV·s}} \approx 5.17\times10^{14}\ \text{Hz}$. …

Why this approach works

The photoelectric effect is governed by Einstein's equation: a photon of energy $h\nu$ must first supply the work function $\phi$ (the minimum energy to free an electron); anything left over becomes the electron's kinetic energy. At the threshold frequency, the photon has just enough energy to free the electron with zero leftover kinetic energy. When a stopping potential $V_0$ is applied, it is just enough to stop even the fastest-ejected electrons, so their maximum kinetic energy equals $eV_0$.

(a) Threshold frequency

At threshold, $K_{\max}=0$, so

Using $\phi=2.14\ \text{eV}$ and $h=4.1357\times10^{-15}\ \text{eV·s}$ (working directly in eV avoids converting $\phi$ to joules):

(b) Wavelength for stopping potential $V_0=0.60\ \text{V}$

The photon's energy must supply both the work function and the stopping energy:

Convert to wavelength using $hc\approx1240\ \text{eV·nm}$:

Consistency check …

Method: Photoelectric Effect Equations

This problem uses Einstein's photoelectric equation, which connects the energy of incident photons to the work function of the metal and the kinetic energy of emitted electrons.

Step 1 — Recall the key relations

The photoelectric equation is:

where:

• $h$ is Planck's constant ($6.63 \times 10^{-34}\ \text{J·s}$)
• $f$ is the frequency of incident light
• $\phi$ is the work function of the metal
• $K_{\text{max}}$ is the maximum kinetic energy of emitted electrons

The stopping potential $V_0$ is related to $K_{\text{max}}$ by:

where $e = 1.6 \times 10^{-19}\ \text{C}$ is the electronic charge.

At the threshold frequency $f_0$, the kinetic energy is zero, so:

Step 2 — Find the threshold frequency (part a)

Given $\phi = 2.14\ \text{eV}$. Convert to joules:

From $hf_0 = \phi$:

Step 3 — Find the wavelength for the given stopping potential (part b)

Given $V_0 = 0.60\ \text{V}$, so:

The incident photon energy is:

Convert to joules: …

Common Mistakes on Photon Energy & Photoelectric Effect Questions

Students lose marks on this exact type of problem in predictable ways. Here are the most frequent errors and how to fix them.

Mistake 1: Forgetting to convert electron-volts to joules

The work function is given as $2.14\ \text{eV}$, but the formula $h f_0 = \phi$ requires $h$ in $\text{J·s}$ and $\phi$ in joules. Many students plug $2.14$ directly into $f_0 = \phi / h$ and get a nonsense answer.

How to avoid: Always convert eV to joules using $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$. So $\phi = 2.14 \times 1.6 \times 10^{-19} = 3.424 \times 10^{-19}\ \text{J}$. Only then divide by $h = 6.63 \times 10^{-34}\ \text{J·s}$.

Mistake 2: Confusing threshold frequency with threshold wavelength

Part (a) asks for threshold frequency $f_0$, not wavelength. Some students compute $\lambda_0 = hc / \phi$ instead. That's the threshold wavelength — a different quantity.

How to avoid: Read the question carefully. Threshold frequency comes from $f_0 = \phi / h$. Threshold wavelength comes from $\lambda_0 = hc / \phi$. They are not interchangeable.

Mistake 3: Using the wrong equation for part (b)

For the stopping potential part, students sometimes write $eV_0 = hf$ alone, forgetting that the work function must be subtracted. The correct relation is:

or equivalently

How to avoid: Remember the photoelectric equation: kinetic energy of the fastest electron equals photon energy minus work function. Stopping potential just measures that maximum kinetic energy: $K_{\text{max}} = eV_0$.

Mistake 4: Mixing up units of stopping potential

The stopping potential is $0.60\ \text{V}$. When you compute $eV_0$, you get $0.60\ \text{eV}$ — that's already in electron-volts. But if you then add it to $\phi$ (which is also in eV), you must keep everything consistent. Some students convert $eV_0$ to joules but leave $\phi$ in eV, or vice versa.

How to avoid: Work entirely in eV for the addition step, then convert to joules only when you need to use $h$ or $c$. Here's the cleanest path:

1. Total photon energy in eV: $E = \phi + eV_0 = 2.14 + 0.60 = 2.74\ \text{eV}$
2. Convert to joules: $E = 2.74 \times 1.6 \times 10^{-19} = 4.384 \times 10^{-19}\ \text{J}$
3. Then $\lambda = hc / E$

Mistake 5: Using $c = 3 \times 10^8$ but forgetting to check units

When computing $\lambda = hc / E$, if $E$ is in joules and $h$ in J·s, $c$ in m/s gives $\lambda$ in metres. Students sometimes leave $E$ in eV and get a wavelength off by $10^{-19}$.

How to avoid: Write out the units explicitly before calculating. If $E$ is in joules, $hc/E$ gives metres. If you prefer to keep $E$ in eV, use the shortcut: …