Question

In photoelectric emission process from a metal of work function $1.6\,eV$ , the kinetic energy of most energetic electrons is $0.4\,eV$ . if the frequency of incident photon is doubled, then corresponding stopping potential will be:

Answer 1

In this above question, we will use the equation of photoelectric effect and we know that the kinetic energy of a photoelectron is = (energy imparted by photon) - (energy used to come out of the surface). And then compare both the obtained equation and hence, solving will get the required solution. When sufficient frequency light is incident on a metal surface, electrons are emitted from the metal surface.

Formula used:
$K.E = h\nu - \phi$
Where, $h$ is Planck's constant and $\nu$ is the frequency.

Complete step by step answer:
Let us assume that ${V_0}$ is the stopping potential.And we know that in a photoelectric effect, the maximum kinetic energy of an electron is,
$K.{E_{\max }} = e{V_s}$
$\Rightarrow {V_s} = \dfrac{{K.{E_{\max }}}}{e}$
Now. Substituting the value in above equation,
$\Rightarrow {V_s} = \dfrac{{0.4eV}}{e} \\\ \Rightarrow \Rightarrow {V_s} = 0.4V \\\$
And we know that the kinetic energy of a photoelectron is = (energy imparted by a photon) - (energy used to come out of the surface). So, energy is constant for a surface, and it is denoted by $\phi$. This is known as the work function of a surface and is always constant for the given material. so, the equation is,
$K.E = h\nu - \phi$...........(Einstein's photoelectric equation)
So, we can write as,
$e{V_s} = h\nu - \phi \\\ \Rightarrow \nu = \dfrac{{e{V_s} + \phi }}{h} \\\$
Now, we have to check the corresponding stopping potential, therefore,
$\dfrac{{{\nu _1}}}{{{\nu _2}}} = \dfrac{{\dfrac{{e{V_{s_1}} + \phi }}{h}}}{{\dfrac{{e{V_{s_2}} + \phi }}{h}}} \\\ \Rightarrow \dfrac{{{\nu _1}}}{{{\nu _2}}} = \dfrac{{e{V_{s_1}} + \phi }}{{e{V_{s_2}} + \phi }} \\\$
And it is given that the frequency of incident photon is doubled therefore,
$\dfrac{{{\nu _1}}}{{{\nu _2}}} = \dfrac{{e{V_{s_1}} + \phi }}{{e{V_{s_2}} + \phi }}$
Now substituting all the values in above equation and calculating,
$\Rightarrow \dfrac{1}{2} = \dfrac{{0.4 + 1.6}}{{e{V_{s_2}} + 1.6}}$
Now, cross multiplying,
$e{V_{s_2}} + 1.6 = 0.8 + 3.2 \\\ \Rightarrow e{V_{s_2}} = 4 - 1.6 \\\ \Rightarrow e{V_{s_2}} = 2.4eV \\\ \therefore {V_{s_2}} = 2.4V \\\$
So, the corresponding stopping potential is $2.4V$.

Note: Note that, in solar panels we can find the photoelectric effect is most common. Photoelectric works on the basic principle of light striking to the cathode which originates the emission of electrons, in turn which produces a current. Photoelectric effect is also used in the form of photo-multiplier tubes.