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 (Threshold condition): The least energetic photon that can still knock out an electron is the one whose energy exactly matches the work function \(W_0\). Any lower frequency carries too little energy and no emission occurs. Thus the cut-off frequency is \(\nu_0 = W_0/h\).
Step 2 (Keep energy in eV, convert \(h\) to eV·s): Using \(h = 6.6\times10^{-34}\ \text{J·s} = \dfrac{6.6\times10^{-34}}{1.6\times10^{-19}} = 4.125\times10^{-15}\ \text{eV·s}\).
Step 3 (Divide work function by this \(h\)):
\[ \nu_0 = \frac{W_0}{h} = \frac{3.3\ \text{eV}}{4.125\times10^{-15}\ \text{eV·s}} \]
Step 4 (Compute):
\[ \nu_0 = 8.0\times10^{14}\ \text{Hz} \]
Same answer, obtained without first converting eV to joule.
\[\boxed{\nu_0 = 8.0\times10^{14}\ \text{Hz}}\]