Question

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV successively illuminate a metal of work function 0.5 eV. The ratio of the maximum speeds of the emitted electrons will be
a. 1:5
b. 1:4
c. 1:2
d. 1:1

Answer 1

Hint: The relation among incident energy of photon, work function and kinetic energy during photoelectric emission is given by the formula,
$E = \phi + {K_{\max }}$

Where,
E = Energy of incident light
$\phi$= work function of metal
${K_{\max }}$= Maximum kinetic energy of emitted electron
${K_{\max }}$ will have the maximum speed electron emitted from the metal plate.

Step by step solution :
Let the maximum speeds of the emitted electrons be ${v_1}$and ${v_2}$ respectively.
According to photoelectric equation,
$E = \phi + \dfrac{1}{2}m{v^2}$ $\Leftrightarrow \dfrac{1}{2}m{v^2} = E - \phi$

For the two emitting electrons the equation are as follows :
$\dfrac{1}{2}mv_1^2 = {E_1} - \phi \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)$

$\dfrac{1}{2}mv_2^2 = {E_2} - {\phi _{}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)$

On dividing (1) by (2) we get
$\dfrac{{v_1^2}}{{v_2^2}} = \dfrac{{{E_1} - \phi }}{{{E_2} - {\phi _{}}}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)$

On substituting the values we get
$\dfrac{{v_1^2}}{{v_2^2}} = \dfrac{{1 - 0.5}}{{2.5 - 0.5}}$

$\dfrac{{{v_1}}}{{{v_{{2_{}}}}}} = \sqrt {\dfrac{1}{4}}$ $\Leftrightarrow \dfrac{{{v_1}}}{{{v_2}}} = \dfrac{1}{2}$

Therefore the ratio of maximum speeds of electrons emitted by two different energies of photons is 1:2.

Hence the correct option is C.

Note: Work function of the metal and mass of the electron always remains constant. Work function is defined as the minimum threshold energy required to emit an electron from a metal. Its unit is electron volt. If the incident frequency of light is less than threshold frequency then whatever the intensity of light the electron will not emit from the metal.