Question

The work function of a metal is 3.3 eV. Calculate the minimum frequency of photon that will emit the photoelectron.
(Given: \( h = 6.6\times10^{-34} \) J·s, \( 1\ \text{eV} = 1.6\times10^{-19} \) J)

Hint:

At the threshold the photon energy just equals the work function: \( h\nu_0 = W_0 \), so \( \nu_0 = W_0/h \). Convert 3.3 eV to joule first.

Answer 1

Step 1 (Concept): Photoelectrons are just emitted when the whole photon energy equals the work function. This minimum (threshold) frequency \( \nu_0 \) satisfies
\[ h\nu_0 = W_0 \quad\Rightarrow\quad \nu_0 = \frac{W_0}{h} \]
Step 2 (Convert the work function to joule):
\[ W_0 = 3.3\ \text{eV} = 3.3\times1.6\times10^{-19} = 5.28\times10^{-19}\ \text{J} \]
Step 3 (Substitute):
\[ \nu_0 = \frac{5.28\times10^{-19}}{6.6\times10^{-34}} \]
Step 4 (Arithmetic):
\[ \nu_0 = 0.8\times10^{15} = 8.0\times10^{14}\ \text{Hz} \]
\[\boxed{\nu_0 = 8.0\times10^{14}\ \text{Hz}}\]