Question

Light of two different frequencies whose photons have
energies 1 eV and 2.5 eV respectively illuminate a metallic
surface whose work function is 0.5 eV successively. Ratio
of maximum speeds emitted electrons will be:​

Answer 1

Answer: Ratio of maximum speeds emitted will be 1:2

Explanation:

For maximum speed of the photo electrons :

$\dfrac{1}{2}mv^2 = E_{photon} - \phi$

where ϕ=0.5 eV is the work function of the metal.

Energy of photon $E_{1}$ = 1 eV

∴ $\dfrac{1}{2}mv_1^2 =1- 0.5$

We get   $\dfrac{1}{2}mv_1^2 =0.5 eV$     .................. equation 1

Energy of photon $E_{2}$ = 2.5 eV

∴  $\dfrac{1}{2}mv_2^2 =2.5- 0.5$

We get   $\dfrac{1}{2}mv_2^2 =2 eV$   ...................equation 2

Ratio of maximum speeds will be  $\frac{v_1^2}{v_2^2}$  = $\frac{0.5}{2}$ = $\frac{1}{4}$

$\dfrac{v_1}{v_2} = \dfrac{1}{2}$

Answer 2

Answer:

Hey Buddy here's ur answer

1:2