#### Question

The work functions for potassium and caesium are 2.25 eV and 2.14 eV respectively. Is the photoelectric effect possible for either of them if the incident wavelength is 5180 Å?

[Given : Planck’s constant = 6.63 x 10^{–34} J.s.;

Velocity of light = 3 x 10^{8} m/s; 1 eV = 1.6 x 10^{–19} J]

#### Solution

Given: (W_{0})_{P} = 2.25 eV = 2.25 x 1.6 x 10^{-19} J = 3.6 x 10^{-19} J,

(W_{0})_{C} = 2.14 eV = 2.14 x 1.6 x 10^{-19} J = 3.424 x 10^{-19} J,

λ = 5180 Å = 5.18 x 10^{-7} m

To find: Will the photoelectric effect occur for either of these elements with λ = 5180 Å

Formula: W_{0} = hv_{0}

`(W_0)_p=h(v_0)_p`

`(v_0)_p=(W_0)_p/h=(3.6xx10^-19)/(6.63xx10^-34)`

`(v_0)_p=5.430xx10^14 Hz`

Similarly

`(v_0)_c=(W_0)_c/h=(3.424xx10^-19)/(6.63xx10^-34)`

`(v_0)_c=5.164 xx10^14Hz`

The corresponding frequency is given by,

`v_2=c/lambda=(3xx10^8)/(5.18xx10^-7)=3/5.18xx10^15`

`v_2=5.792xx10^14Hz`

for potassium

`5.792xx10^14Hz>5.430xx10^14Hz`

`i.e, v_2>(v_o)_p`

Photoelectric emission will take place when light of wavelength λ is incident on it. For caesium,

`5.792xx10^14Hz>5.16xx10^14Hz`

`i.e, v_2>(v_o)_c`

Photoelectric emission will take place when light of wavelength λ is incident on it. For λ= 5180Å wavelength, both potassium and caesium will exhibit photoelectric emission.