Question

Light of two different frequencies whose photons have energies $1eV$ and $2.5eV$ respectively, successively illuminate a metallic surface whose work function is $0.5eV$. Ratio of maximum speeds of emitted electrons will be:
(A). $1:4$
(B). $1:1$
(C). $1:5$
(D). $1:2$

Answer 1

When the photons strike the metal, their energy is transferred to the electrons. If the energy possessed by the photons is greater than work function then the electrons leave the metal surface. The energy of photons is equal to the sum of kinetic energy of electron and work function. Using the following relation, we can calculate the kinetic energies and find the ratio.
Formulas used:
$h\nu =W+K$

Complete step-by-step solution:
When light strikes a metal, the energy possessed by its photons is transferred to the electrons. The minimum energy required by an electron to leave the surface is known as the work function. Therefore,
$h\nu =W+K$ ---------- (1)
Here, $h\nu$ is the energy possessed by the photons of light and $\nu$ is the frequency
$W$ is the metal’s work function
$K$ is the kinetic energy
Given, $h{{\nu }_{1}}=1eV$ and $h{{\nu }_{2}}=2.5eV$, $W=0.5eV$
From eq (1), the kinetic energy of the electrons is given as-
$K=h\nu -W$
Substituting given values for first frequency, we get,

& {{K}_{1}}=h{{\nu }_{1}}-W \\\ & \Rightarrow {{K}_{1}}=(1-0.5)eV \\\ & \therefore {{K}_{1}}=0.5eV \\\ \end{aligned}$$ Therefore, the kinetic energy of electrons for light with frequency$${{\nu }_{1}}$$ is $$0.5eV$$. Substituting given values for second frequency, we get, $$\begin{aligned} & {{K}_{2}}=h{{\nu }_{2}}-W \\\ & \Rightarrow {{K}_{2}}=(2.5-0.5)eV \\\ & \therefore {{K}_{2}}=2eV \\\ \end{aligned}$$ The kinetic energy of electrons for light with frequency $${{\nu }_{2}}$$ is $$2eV$$. The ratio of the kinetic energies of the electrons will be $$\dfrac{{{K}_{1}}}{{{K}_{2}}}$$. Therefore, $$\dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{0.5}{2}$$ $$\therefore \dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{1}{4}$$ ------- (2) We know that kinetic energy is given by- $$K=\dfrac{1}{2}m{{v}^{2}}$$ -------- (3) Here, $$m$$ is the mass of body $$v$$ is the velocity From eq (2) and eq (3), we get, $$\begin{aligned} & \dfrac{{{K}_{1}}}{{{K}_{2}}}=\dfrac{\dfrac{1}{2}mv_{1}^{2}}{\dfrac{1}{2}mv_{2}^{2}}=\dfrac{1}{4} \\\ & \Rightarrow \dfrac{v_{1}^{2}}{v_{2}^{2}}=\dfrac{1}{4} \\\ & \therefore \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{1}{2} \\\ \end{aligned}$$ Therefore, the ratio of velocities of electrons is $$\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{1}{2}$$ **Therefore, the ratio of the velocities of electrons is $$\dfrac{1}{2}$$ . Hence, the correct option is (D).** **Note:** The energy of the photons is directly proportional to the frequency but inversely proportional to the wavelength. The potential difference is applied to accelerate the electrons with kinetic energy. The electrons at the surface have greater kinetic energies than the electrons below the surface.