What is the value of the work function for a metal surface if the threshold frequency is $5\times 10^{14}\text{ Hz}$? (Take $h=6.63\times 10^{-34}\text{ J s}$)

Solution

1. Identify the given values: Planck's constant $h=6.63\times 10^{-34}\text{ J s}$ and threshold frequency ${\nu }_0=5\times 10^{14}\text{ Hz}$.
2. Use the formula for the work function: $\Phi =h{\nu }_0$.
3. Substitute the values into the equation: $$\begin{array}{rl}\Phi & =(6.63\times 10^{-34}\text{ J s})\times (5\times 10^{14}\text{ Hz})\\ & =3.315\times 10^{-19}\text{ J}\end{array}$$.
4. Hence the answer is (A).