Question

The work function of metals $A$ and $B$ are in the ratio $1:2$. If light of frequencies $f$ and $2f$ are incident on metal surfaces $A$and $B$ respectively, the ratio of the maximum kinetic energies of the photoelectrons emitted is :
(A). 1:1
(B). 1:2
(C). 1:3
(D). 1:4

Answer 1

Hint: The factors which affect the maximum kinetic energy of the photoelectrons are the type of the material on the surface on which the light is getting striked and the frequency of the incident radiation. The energy of the electron simply increases with the increasing frequency of the incident radiation on the material’s surface.

Formula Used
${E_{\max }} = h\upsilon - \phi $, where $\upsilon = $ frequency of the incident light and $\phi $= photoelectric work function of the metal.

Complete step-by-step answer:
So now we will apply this formula for the two given metals separately,
So, for the metal $A$,
${E_{\max (A)}} = hf - {\phi _A}$------equation (1), (since given frequency of the incident light for metal $A$=$f$)
And similarly, for the metal $B$,
${E_{\max (B)}} = h(2f) - {\phi _B}$--------equation (2), (since given frequency of the incident light for metal $A$=$2f$)
Now we have got the separate equation for the maximum kinetic energy of a photo electron for both the metals, so now we will be taking the ratio of both equation (1) and equation (2), we get
$\dfrac{{{E_{\max (A)}}}}{{{E_{\max (B)}}}} = \dfrac{{hf - {\phi _A}}}{{h(2f) - {\phi _B}}}$
$ \Rightarrow \dfrac{{{E_{\max (A)}}}}{{{E_{\max (B)}}}} = \dfrac{{\dfrac{{hf}}{{{\phi _A}}} - 1}}{{\dfrac{{2hf}}{{{\phi _A}}} - \dfrac{{{\phi _B}}}{{{\phi _A}}}}}$
$ \Rightarrow \dfrac{{{E_{\max (A)}}}}{{{E_{\max (B)}}}} = \dfrac{{\dfrac{{hf}}{{{\phi _A}}} - 1}}{{\dfrac{{2hf}}{{{\phi _A}}} - 2}}$ ---since ($\dfrac{{{\phi _A}}}{{{\phi _B}}} = \dfrac{1}{2}$)
$ \Rightarrow \dfrac{{{E_{\max (A)}}}}{{{E_{\max (B)}}}} = \dfrac{{\dfrac{{hf - {\phi _A}}}{{{\phi _A}}}}}{{\dfrac{{2hf - 2{\phi _A}}}{{{\phi _A}}}}}$
$ \Rightarrow \dfrac{{{E_{\max (A)}}}}{{{E_{\max (B)}}}} = \dfrac{{(hf - {\phi _A})}}{{2(hf - {\phi _A})}}$
$ \Rightarrow \dfrac{{{E_{\max (A)}}}}{{{E_{\max (B)}}}} = \dfrac{1}{2}$
Hence, the ratio of the maximum kinetic energies of the photoelectrons emitted is $\dfrac{1}{2}$.
So, option (B) is the correct answer.

Note: in the above phenomenon when the light strikes the metal surface then the photo electrons are getting removed from the surface of the metal. Here we see that the role of the work function of the metal came into picture. The work function is the minimum thermodynamic work that is needed to eject an electron from a metal solid surface to a point in a vacuum immediately when light strikes the metal surface.