Question

If the work function of a certain metal is $1.8eV$.(a) What is the stopping potential for electrons ejected from metal when light of $4000{{A}^{0}}$shines on the metal. (b) What is the maximum speed of the ejected electrons?

Answer 1

In this type of questions we have to apply photoelectric equation as some data is given so other parameters can also be identified easily and for velocity we can calculate it from the value of kinetic energy that can be obtained by photoelectric equation and in the same way stopping potential is also calculated.

Complete step by step solution:
Here in this question it is saying that when light of $4000{{A}^{0}}$falls on the metal surface then electrons are ejected with maximum kinetic energy or with maximum velocity.
So first we calculate the energy of incident light as its wavelength is given.
Let us assume its wavelength is represented by $\lambda$.
So we can write
$\lambda =4000{{A}^{0}}$
Energy of incident light is calculated by
$E=\dfrac{hc}{\lambda }$
Where
$h=6.63\times {{10}^{-34}}(S.I.unit)$
$c=3\times {{10}^{8}}\dfrac{m}{s}$
Put this value we can get the energy of incident light,
$E=\dfrac{6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4000\times {{10}^{-10}}}$
$\therefore E=4.9725\times {{10}^{-19}}Joule$
Convert Joule to eV.

& E=\dfrac{4.9725\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV \\\ & \therefore E=3.10eV \\\ \end{aligned}$$ So with this energy light is incident on the metal surface. In this energy some is used in work function and the rest becomes the maximum kinetic energy of emitted photons. Since Work function of metal is given in the question as $$1.8eV$$. Apply Photoelectric equation we get , $$E={{W}_{0}}+K.E{{.}_{\max }}$$ Where $${{W}_{0}}=Work \, finction$$ Put the value in this equation we get, $$\begin{aligned} & 3.10=1.8+K.E{{.}_{\max }} \\\ & \Rightarrow K.E{{.}_{\max }}=3.10-1.8 \\\ & \therefore K.E{{.}_{\max }}=1.3eV \\\ \end{aligned}$$ This is the required maximum kinetic energy of emitted photons. (i) In this part we have to find stopping potential . Let us assume the stopping potential can be represented as $${{V}_{0}}$$. Since relation between kinetic energy and stopping potential is $$K.E{{.}_{\max }}=e{{V}_{0}}$$ Put the value of Kinetic energy we get , $$\begin{aligned} & 1.3eV=e{{V}_{o}} \\\ & \therefore {{V}_{0}}=1.3V \\\ \end{aligned}$$ So the stopping potential is $$1.3volt$$. (ii) In second part we have to find the maximum speed of ejected electrons, $$K.E{{.}_{\max }}=\dfrac{1}{2}{{m}_{e}}{{v}_{\max }}^{2}$$ Put the value of Kinetic energy and mass of electrons we get the value of maximum velocity of ejected electrons. First convert kinetic energy into joules so we can write as , $$\begin{aligned} & 1.3eV=1.3\times 1.6\times {{10}^{-19}}J \\\ & \therefore 1.3eV=2.08\times {{10}^{-19}}J \\\ \end{aligned}$$ Put the values in above equation we get, $$2.08\times {{10}^{-19}}=\dfrac{1}{2}\times 9.1\times {{10}^{-31}}\times {{v}_{\max }}^{2}$$ On simplifying , we get $$\begin{aligned} & 4.16\times {{10}^{-19}}=9.1\times {{10}^{-31}}\times {{v}_{\max }}^{2} \\\ & \Rightarrow {{v}_{\max }}^{2}=\dfrac{4.16\times {{10}^{-19}}}{9.1\times {{10}^{-31}}} \\\ & \Rightarrow {{v}_{\max }}=\sqrt{0.457\times {{10}^{12}}} \\\ & \therefore {{v}_{\max }}=0.67\times {{10}^{6}}\dfrac{m}{s} \\\ \end{aligned}$$ So we can conclude that stopping potential is $$1.3volt$$ and maximum velocity of ejected electrons is $$0.67\times {{10}^{6}}\dfrac{m}{s}$$. **Note:** Here stopping potential means at that value of potential difference the ejection of electrons from the surface of metals stops and no current is produced . In photoelectric effect if the energy of incident light is not greater than the work function of the metal then electrons cannot be emitted no current is produced.