Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.

The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from *λ*_{1} to *λ*_{2}. Derive the expressions for the threshold wavelength *λ*_{0} and work function for the metal surface.

State two important features of Einstein's photoelectric equation.

#### Solution

Einstein's photoelectric equations is given by

`K_max=1/2mv_max^2=hv=phi_0`

where

K_{max} = Maximum kinetic energy of the photoelectron

v_{max}_{ }= Maximum velocity of the emitted photoelectron

m = Mass of the photoelectron

ν = Frequency of the light radiation

ϕ_{0}_{ }= Work function

If *ν*_{0} is the threshold frequency, then the work function can be written as

*ϕ*_{0} = h*ν*_{0}

`=>K_max=1/2mv_max^2=hv-hv_0=h(v-v_0)`

The above equations explains the following results:

1. If *ν < ν*_{0}, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if *ν > **ν*_{0}.

2. The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectron depends only on the frequency of incident light.

According to the photoelectric equation,

`K_max=1/2mv_max^2=hv-phi_0`

`K_max=(hc)/lambda_1-phi_0`

Let the maximum kinetic energy for the wavelength of the incident *λ*_{2} be *K*'_{max}

`K'_max=(hc)/lambda_2-phi_0`

Form (i) and (ii), we have

`(hc)/lambda_2-phi_0=2((hc)/lambda_1-phi_0)`

`=>phi_0=hc(2/lambda_1-1/lambda_2)`

`=>hv_0=hc(2/lambda_1-1/lambda_2)`

`=>c/(lambda_0)=c(2/lambda_1-1/lambda_2)`

`=>1/lambda_0=(2/lambda_1-1/lambda_2)`

`=>lambda_0=((lambda_1lambda_2)/(2lambda_2-lambda_1))`