The above reprt the graph electron energy against the frequency of the radiation incident on a metal surface. lnterprete the; (i) slope of the graph; (ii) i...

Answer Details

The graph is a straight line of the maximum kinetic energy \(E\) of the emitted photoelectrons (vertical axis, in joules) against the frequency \(f\) of the incident radiation (horizontal axis, in hertz). The line crosses the frequency axis at the point \(K\) and, when produced backwards, cuts the energy axis below the origin at the point \(C\). The graph is a plot of Einstein's photoelectric equation,

\[ E = hf - hf_0 \]

where \(h\) is Planck's constant and \(f_0\) is the threshold frequency of the metal. Comparing this with the equation of a straight line \(y = mx + c\): \(m = h\) is the gradient and \(c = -hf_0\) is the intercept on the energy axis.

(i) The slope of the graph

Comparing \(E = hf - hf_0\) with \(y = mx + c\), the gradient is

\[ \text{slope} = \frac{\Delta E}{\Delta f} = h \]

So the slope represents Planck's constant, \(h\) \(\left(\approx 6.6\times10^{-34}\,\text{J s}\right)\). It has the same value for every metal.

(ii) The intercept OC

OC is the intercept on the energy axis, obtained when \(f = 0\). From the equation, \(E = -hf_0\), so this intercept is negative and equal in magnitude to the work function \(\varphi\) of the metal:

\[ OC = -hf_0 = -\varphi \]

Thus \(|OC|\) is the minimum energy needed to release an electron from the surface of the metal, i.e. the work function.

(iii) The intercept OK

OK is the intercept on the frequency axis, obtained when the photoelectron energy \(E = 0\). Setting \(0 = hf - hf_0\) gives \(f = f_0\). Therefore OK represents the threshold (cut-off) frequency \(f_0\), the minimum frequency of the incident radiation below which no electrons are emitted from the metal.