Q.The work function for a certain metal is $4.2\ \text{eV}$. Will this metal give photoelectric emission for incident radiation of wavelength $330\ \text{nm}$?

Concept: Maximum Kinetic Energy — photoelectric emission occurs only if the incident photon energy exceeds the work function.

Step 1: Find the photon energy.

Wavelength $\lambda = 330\ \text{nm} = 330 \times 10^{-9}\ \text{m}$.

Photon energy $E = \frac{hc}{\lambda}$. Using $hc = 1240\ \text{eV·nm}$:

Step 2: Compare with work function $\phi = 4.2\ \text{eV}$. …

Why this comparison works

Photoelectric emission happens only when an incident photon has enough energy to overcome the binding energy holding an electron in the metal. That minimum required energy is the work function ($\phi$). If the photon's energy ($E$) is less than $\phi$, the electron simply cannot be freed — no matter how many photons hit the surface.

So the entire problem reduces to one question: Is the photon energy from a $330\ \text{nm}$ wave greater than or equal to $4.2\ \text{eV}$?

Step-by-step solution

1. Find the photon energy in joules first. The energy of a single photon is given by $E = \frac{hc}{\lambda}$, where:
   • $h = 6.63 \times 10^{-34}\ \text{J·s}$ (Planck's constant)
   • $c = 3.00 \times 10^8\ \text{m/s}$ (speed of light)
   • $\lambda = 330\ \text{nm} = 330 \times 10^{-9}\ \text{m}$

Compute step by step:

2. Convert this energy into electronvolts. Since $1\ \text{eV} = 1.602 \times 10^{-19}\ \text{J}$, we divide:

3. Compare with the work function. …

Method: Photoelectric Effect Threshold Condition

We check whether the incident photon has enough energy to overcome the metal's work function. If the photon energy is greater than or equal to the work function, emission occurs.

Step 1: Convert work function to joules (or keep in eV — we'll compare in eV).

Work function $\phi = 4.2\ \text{eV}$.

Step 2: Find the energy of the incident photon.

Photon energy $E = \frac{hc}{\lambda}$, where $h = 6.63 \times 10^{-34}\ \text{J·s}$, $c = 3 \times 10^8\ \text{m/s}$, and $\lambda = 330\ \text{nm} = 330 \times 10^{-9}\ \text{m}$.

First compute in joules:

Convert to eV (1 eV = $1.6 \times 10^{-19}\ \text{J}$):

Step 3: Compare photon energy with work function.

Photon energy $= 3.77\ \text{eV}$

Work function $= 4.2\ \text{eV}$ …

The most common mistake here is rushing to compare the work function directly with the photon energy without checking units. The work function is given in eV, but the wavelength is in nm — you must convert everything to a consistent unit system before comparing.

Mistake 1: Forgetting to convert wavelength to energy in eV

Students often calculate the photon energy in joules and then compare it directly to 4.2 eV without converting. That gives a meaningless comparison.

How to avoid: Always compute the photon energy in the same unit as the work function. Use the formula

with $h = 6.63 \times 10^{-34}\ \text{J·s}$, $c = 3 \times 10^8\ \text{m/s}$, and $\lambda = 330\ \text{nm} = 330 \times 10^{-9}\ \text{m}$.

First get $E$ in joules:

Now convert to eV using $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$:

Mistake 2: Comparing the wrong quantities

Some students compare the photon energy to the threshold frequency or to the maximum kinetic energy instead of the work function. The condition for emission is simple: photoelectric emission occurs only if the incident photon energy $E$ is greater than or equal to the work function $\phi$.

How to avoid: State the condition clearly before plugging numbers. Write:

For emission: $E \ge \phi$

Here $E = 3.77\ \text{eV}$ and $\phi = 4.2\ \text{eV}$. Since $3.77 < 4.2$, emission does not occur.

Mistake 3: Confusing wavelength and frequency

A student might compute the frequency from $\lambda$ and then compare it to the threshold frequency. That's fine in principle, but it adds an extra step where unit errors can creep in. The threshold frequency is $f_0 = \phi/h$, and you'd need to check if $f > f_0$. It's safer to work directly with energy.

How to avoid: Stick to one method. The energy comparison is the most direct: compute $E$ in eV, compare to $\phi$ in eV.

Mistake 4: Misinterpreting "work function" …