Question

The work functions of Silver and Sodium are 4.6 and 2.3 eV respectively. Find out the ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium.

Answer 1

Hint : By the term work function, we mean the minimum amount of energy that will be required in order to remove an electron that is present on the surface of a solid to any point present on the vacuum, at that every point which will mark a space outside that of the solid surface.

Complete step by step answer:
Let us first find the relation between the expression of the stopping potential. So, the expression is given as:
$\mathrm{he}=\mathrm{hv}_{\mathrm{o}}+\mathrm{KE}$
In the above expression,
h is defined as the Planck’s constant
e is the energy of one electron
${{v}_{0}}$is the velocity at which the photon will eject itself from the surface of the metal.
And KE is the kinetic energy.
Now, we can write the expression as $=\mathrm{hv}_{\mathrm{o}}+\mathrm{eV}_{\mathrm{o}}$
Here, $\mathrm{V}_{\mathrm{o}}$ =stopping potential
So, if we form the expression with respect to stopping potential we will get:
$\mathrm{V}_{\mathrm{o}}=\mathrm{h} / \mathrm{e}\left(\mathrm{v}-\mathrm{v}_{\mathrm{o}}\right)$
This can also be written as:
$\mathrm{V}_{\mathrm{o}}=(\mathrm{h} / \mathrm{e}) \mathrm{v}-(\mathrm{h} / \mathrm{e}) \mathrm{v}_{\mathrm{o}}$
The above developed expression is in the form of: $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
Thus, we can say that:
The slope or the m will be: $\mathrm{m}=\mathrm{h} / \mathrm{e}$
The y-intercept will be given as: $-\mathrm{hv}_{\mathrm{o}} / \mathrm{e}$
Thus, from the slope of this graph we can conclude the value as: $\mathrm{h} / \mathrm{e}$.
Therefore, it can be proved that the slope of the graph does not depend on the work function. Hence, the ratio has to be 1.

Note: It should also be known that at a very higher intensity, the radiation will produce a higher value of the photo current. If we want to achieve a negative potential value then in that case the absolute value of the potential difference will increase, and the value of the photo current will fall down and reach zero at the stopping potential.