Question

When photons of energy 4.25eV strike the surface of a metal A, the ejected photoelectrons have the maximum kinetic energy of \[{K_A}{\text{ }}eV\] and de Broglie wavelength ${\lambda _A}{\text{ }}$. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70eV is ${K_B} = \left( {{K_A} - 1.5} \right)eV$. If the de Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$ then
(This question has multiple correct options)
(a) The work function of A is 2.25eV
(b) The work function of B is 4.20eV
(c) \[{K_A} = 2.00eV\]
(d) ${K_B} = 2.75eV$

Answer 1

Hint: In this question use the relationship between the Kinetic energy and the momentum that is $K.E = \dfrac{1}{2}\dfrac{{{P^2}}}{m}$. Find the dependency of the kinetic energy over the wavelength of the particle, use the constraints of the questions to get to the right options.

Complete step-by-step solution -
As we all know, the kinetic energy (K.E) of any particle is half times the mass (m) of the particle multiplied by the square of the velocity (v).
$ \Rightarrow K.E = \dfrac{1}{2}m{v^2}$................... (1)
And we all know that momentum (P) of any particle is the product of mass (m) and velocity (v) of the particle.
$ \Rightarrow P = mv$..................... (2)
Now from equation (1) and (2) we have,
$ \Rightarrow K.E = \dfrac{1}{2}\dfrac{{{P^2}}}{m}$
So kinetic energy is directly proportional to the square of the momentum.
$ \Rightarrow K \propto {P^2}$................ (1)
Now as we all know that wavelength $\left( \lambda \right)$ is inversely proportional to the momentum (P).
$ \Rightarrow \lambda \propto \dfrac{1}{P}$.............. (2)
Now from equation (1) and (2) we have,
$ \Rightarrow K \propto \dfrac{1}{{{\lambda ^2}}}$
Now there are two metals given A and B so the ratio of the kinetic energy of A and B in terms of the wavelength ${\lambda _A}{\text{ and }}{\lambda _B}$ are
\[ \Rightarrow \dfrac{{{K_A}}}{{{K_B}}} = \dfrac{{\lambda _B^2}}{{\lambda _A^2}}\]
Now it is given that ${\lambda _B} = 2{\lambda _A}$ so use this property in above equation we have,
\[ \Rightarrow \dfrac{{{K_A}}}{{{K_B}}} = \dfrac{{{{\left( {2{\lambda _A}} \right)}^2}}}{{\lambda _A^2}} = 4\]
$ \Rightarrow {K_A} = 4{K_B}$................... (3)
Now it is also given that ${K_B} = \left( {{K_A} - 1.5} \right)eV$
So from equation (3) substitute the value of ${K_A}$ in the above equation we have,
$ \Rightarrow {K_B} = \left( {4{K_B} - 1.5} \right)eV$
Now simplify this we have,
$ \Rightarrow 3{K_B} = \left( {1.5} \right)eV$
$ \Rightarrow {K_B} = \dfrac{{1.5}}{3} = 0.5eV$
Now from equation (3) we have,
$ \Rightarrow {K_A} = 4\left( {0.5} \right) = 2eV$
Now as we know that the work function is the difference of the energy of photons by which metal strikes and the kinetic energy of the metal.
Now it is given that photons of energy 4.25eV strike the surface of the metal A and photons with energy 4.70eV strike the surface of the metal B.
So the work function of the metal A is = 4.25eV - ${K_A}$ = 4.25 – 2 = 2.25eV.
And the work function of the metal B is = 4.70eV - ${K_B}$ = 4.70 – 0.5 = 4.2eV.
So this is the required answer.

Hence option (A), (B) and (C) are the correct options.

Note: The basic definition of work function is quite useful while dealing with problems of such kind. As metals are composed of electrons so when energy is incident over it some electrons are emitted out so this minimum energy of the light that if incident over a metal surface removes an electron to infinity is called as the work function of that metal.