Question

The work function for a metal having threshold frequency $5\times 10^7\text{ Hz}$ is?

Hint:

Memorizing the approximate value of Planck's constant $h \approx 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ is essential for modern physics problems.
If a question ever requires the final answer in electron-volts (eV), divide the calculated Joules by $1.6 \times 10^{-19}$.

Answer 1

Step 1: Understanding the Concept:
The work function of a metal describes the minimum amount of energy required to eject an electron completely from its surface.
This energy is directly proportional to the threshold frequency, which is the absolute minimum frequency of incident light needed to trigger photoelectric emission.
Step 2: Key Formula or Approach:
The work function $\Phi$ is calculated using the equation $\Phi = h \nu_0$.
Here, $h$ represents Planck's constant ($6.626 \times 10^{-34}\text{ J}\cdot\text{s}$) and $\nu_0$ is the given threshold frequency.
Step 3: Detailed Explanation:
The problem states the threshold frequency is $\nu_0 = 5 \times 10^7\text{ Hz}$.
Substituting this value into the work function formula yields:
\[ \Phi = (6.626 \times 10^{-34}\text{ J}\cdot\text{s}) \times (5 \times 10^7\text{ Hz}) \] \[ \Phi = 33.13 \times 10^{-27}\text{ J} \] Adjusting the decimal place to express the answer in standard scientific notation:
\[ \Phi = 3.313 \times 10^{-26}\text{ J} \] Step 4: Final Answer:
The work function of the metal is approximately $3.31 \times 10^{-26}\text{ J}$.