Q.Light of frequency $7.21 \times 10^{14}\ \text{Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^{5}\ \text{m/s}$ are ejected from the surface. What is the threshold frequency for photoemission of electrons?

Concept: Maximum Kinetic Energy — the photoelectric equation relates the incident photon energy to the work function and the maximum kinetic energy of the ejected electron.

The photoelectric equation is:

where $\phi = h f_0$ is the work function and $f_0$ is the threshold frequency.

Step 1: Write the equation in terms of $f_0$:

Step 2: Solve for $f_0$:

Step 3: Substitute values. Use $m = 9.1 \times 10^{-31}\ \text{kg}$, $h = 6.63 \times 10^{-34}\ \text{J·s}$:

$f_0 = 7.21 \times 10^{14} - \frac{(9.1 \times 10^{-31})(6.0 \times 10^{5})^2}{2 \times 6.63 \times 10^{-34}}$ …

The core idea here is the photoelectric effect: when light hits a metal, each photon gives its energy $hf$ to an electron. The electron uses some of that energy to escape the metal (the work function $\phi = hf_0$), and the rest becomes kinetic energy. The maximum kinetic energy occurs for electrons that escape without losing energy to collisions inside the metal.

So the equation is:

where $f_0$ is the threshold frequency — the minimum frequency needed to eject any electron at all.

We know $f = 7.21 \times 10^{14}\ \text{Hz}$ and the maximum speed $v_{\text{max}} = 6.0 \times 10^{5}\ \text{m/s}$. We need $f_0$.

1. Find the maximum kinetic energy. The kinetic energy is $K_{\text{max}} = \frac{1}{2} m_e v_{\text{max}}^2$, where $m_e = 9.11 \times 10^{-31}\ \text{kg}$ (electron mass).

First square the speed: $(6.0 \times 10^{5})^2 = 3.6 \times 10^{11}$.

Then multiply: $9.11 \times 10^{-31} \times 3.6 \times 10^{11} = 3.2796 \times 10^{-19}$.

Half of that: $K_{\text{max}} = 1.6398 \times 10^{-19}\ \text{J}$.

2. Write the photoelectric equation.

So

and

3. Plug in the numbers. Planck's constant $h = 6.626 \times 10^{-34}\ \text{J·s}$. First compute $hf$:

(Check: $6.626 \times 7.21 \approx 47.77$, and $10^{-34} \times 10^{14} = 10^{-20}$, so $4.777 \times 10^{-19}$ — correct.) …

Method: Photoelectric Equation Approach

This problem uses Einstein's photoelectric equation, which connects the incident photon energy, the work function (or threshold frequency), and the maximum kinetic energy of ejected electrons.

Step 1 – Write the photoelectric equation

The maximum kinetic energy of ejected electrons is given by:

where $h$ is Planck's constant, $f$ is the incident frequency, and $f_0$ is the threshold frequency.

Step 2 – Express $K_{\text{max}}$ in terms of the given speed

The maximum kinetic energy is also:

where $m$ is the electron mass ($9.1 \times 10^{-31}\ \text{kg}$) and $v_{\text{max}} = 6.0 \times 10^{5}\ \text{m/s}$.

Step 3 – Equate and solve for $f_0$

From the two expressions:

Rearranging for $f_0$:

Step 4 – Substitute values

Take $h = 6.63 \times 10^{-34}\ \text{J·s}$, $m = 9.1 \times 10^{-31}\ \text{kg}$, $f = 7.21 \times 10^{14}\ \text{Hz}$, and $v_{\text{max}} = 6.0 \times 10^{5}\ \text{m/s}$.

First compute the kinetic energy term:

$\frac{m v_{\text{max}}^2}{2} = \frac{(9.1 \times 10^{-31})(6.0 \times 10^{5})^2}{2}$ …

Students often lose marks on this question not because the photoelectric equation is hard, but because they rush or mis-handle units and constants. Here are the most common mistakes and how to avoid each.

Mistake 1: Forgetting to convert electron volts to joules (or vice versa)

The photoelectric equation $K_{\text{max}} = hf - \phi$ uses $h = 6.63 \times 10^{-34}\ \text{J·s}$. If you try to work in eV without converting consistently, you'll get a wrong numerical answer. Many students compute $hf$ in joules, then subtract a work function in eV — that's mixing units.

How to avoid: Stick entirely to SI units (joules, kg, m/s) throughout the calculation. Only convert to eV at the very end if the question asks for it. Here, the threshold frequency is asked in Hz, so stay in joules.

Mistake 2: Using the wrong expression for kinetic energy

The maximum kinetic energy of ejected electrons is $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$, where $m$ is the electron mass ($9.11 \times 10^{-31}\ \text{kg}$). Some students mistakenly use $mv$ (momentum) or forget the $\frac{1}{2}$ factor.

How to avoid: Write the kinetic energy formula explicitly before plugging numbers. Double-check that you've squared the speed and multiplied by half.

Mistake 3: Confusing threshold frequency with threshold wavelength

The threshold frequency $f_0$ is related to the work function by $\phi = h f_0$. Some students try to use $\lambda_0 = c / f_0$ prematurely, or they solve for wavelength when the question asks for frequency.

How to avoid: Read the question carefully — it asks for threshold frequency. Solve directly from $f_0 = \frac{\phi}{h}$ after finding $\phi$. Don't introduce wavelength unless needed.

Mistake 4: Arithmetic or exponent errors with large/small numbers

The numbers here are typical: $h \approx 6.63 \times 10^{-34}$, $m_e \approx 9.11 \times 10^{-31}$, speeds around $10^5$–$10^6$ m/s, frequencies around $10^{14}$ Hz. A single exponent slip (e.g., writing $10^{-20}$ instead of $10^{-19}$) changes the answer completely.

How to avoid: Work step by step, writing each intermediate result in scientific notation. Use your calculator carefully — enter the full expression at once if possible, or check the exponent after each multiplication.

Mistake 5: Forgetting that $K_{\text{max}}$ is the maximum kinetic energy …