Question

Light of two different frequencies whose photons have energies 1.3 eV and 2.8 eV respectively, successfully illuminate a metallic surface whose work function is 0.8 eV. The ratio of maximum speeds of emitted electrons will be ______.

Options

• 1 : 3

• 1 : 2

• 1 : 5

• 1 : 4

Answer 1

Light of two different frequencies whose photons have energies 1.3 eV and 2.8 eV respectively, successfully illuminate a metallic surface whose work function is 0.8 eV. The ratio of maximum speeds of emitted electrons will be 1 : 2.

Explanation:

The energy of the photon, `E_"photon" = KE + Φ`

where, KE = kinetic energy

and Φ = work function.

The energy of photon 1, `E_{"photon"_1}` = 1.3 eV

`E_{"photon"_1} = 1/2m"v"^2 + 0.8`

`1/2m"v"_1^2 = 1.3 - 0.8 ⇒ 1/2m"v"_1^2 = 0.5`

Energy of photon 2, `E_{"photon"_2} = 2.8` eV

∴ `1/2m"v"_2^2 = 2.8 - 0.8 = 2` eV

Now, ratio of maximum speed = `("v"_1^2)/("v"_2^2) = 0.5/2 = 1/4`

∴ `"v"_1/"v"_2 = 1/2`