Question

When a certain photosensitive material is illuminated by monochromatic light of frequency f, the stopping potential for the photocurrent is v / 2. When the surface is illuminated by the monochromatic light of frequency f / 2 the stopping potential is v. The threshold frequency for the photoelectric emission is?

Answer 1

Given :

• When a photosensitive material is illuminated by monochromatic light of frequency f, stopping potential is V/ 2.
• When it's illuminated by the monochromatic light of frequency f / 2 the stopping potential is v..

To Find :

The threshold frequency for the photoelectric emission

Theory:

• Einstein's Photoelectric Equation :

$\rm{hf=W+KE_{max}}$

and $\rm\:KE_{max}=eV_s$

Then,

$\bf\blue{hf=W+eV_s}$

Where,

• h = plank constant
• W = Work function
• $\sf\:eV_{s}=$ Stopping potential
• f = frequency

• Work Function :

The minimum energy required for emission of electrons from metal is called work function.

$\sf\:W=hf_o$

Where, $\sf\:f_o$ is the threshold frequency.

Solution :

Let the threshold frequency for the photoelectric emission be $\sf\:f_o$ and the work function of photosensitive material be W

Case -1 :

• When Frequency = f and
• Stopping potential = V/2

Then, by Einstin's Photoelectric Equation:

$\sf{hf=W+\dfrac{V}{2}}$ Equation (1)

Case -2 :

• When Frequency = f/2
• Stopping potential = V

Then, by Einstin's Photoelectric Equation:

$\sf{h\times\dfrac{f}{2}=W+V}$ Equation (2)

Now , Subtract equation (2) from (1) , then

$\sf\implies{f-\dfrac{f}{2}=\dfrac{V}{2}-V}$

$\sf\implies\dfrac{2f-f}{2}=\dfrac{V-2V}{2}$

$\sf\implies\:V=-hf$

Put V = - hf in equation (2) , then

$\sf\implies\dfrac{hf}{2}=W-hf$

$\sf\implies\dfrac{hf}{2}=hf_{o}-hf$

$\sf\implies\dfrac{hf}{2}+hf=hf_o$

$\sf\implies\dfrac{hf+2hf}{2}=hf_o$

$\sf\implies\dfrac{3hf}{2}=hf_o$

$\sf\implies\:f_o=\dfrac{3f}{2}$

Therefore ,The threshold frequency for the photoelectric emission is 3f/2 .

Answer 2

Given:-

• When a certain photosensitive material is illuminated by monochromatic light of frequency f, the stopping potential for the photocurrent is v/2.When the surface is illuminated by the monochromatic light of frequency f/2 the stopping potential is V .

To Find:-

• The threshold frequency for the photoelectric emission = ?

Solution:-

Let the threshold frequency for the Photoelectric emission be $\sf{f_o}$ and the work function of Photosensitive material be W.

Case-1 :-

• When frequency is = f and
• stopping potential = v/2

Then, by Einstein's Photoelectric Equation:

$\sf{hf\:=\:W + V/2}$ Equation (1)

Case -2 :-

• When frequency is = f/2
• Stopping potential = V

Then,by Einstein's Photoelectric Equation:

$\sf{h × f/2\:=\:W+ V}$ Equation (2)

Now, Subtract Equation (2) from (1) ,then

$\sf\implies{f \: –f/2\:=\:V/2 - V}$

$\sf\implies{ 2f -f /2\:=\: V-2/2}$

$\sf\implies{V \:= \: –hf}$

Put V = - hf in equation (2),then

$\sf\implies{\frac{hf}{2}\:=\:W - hf}$

$\sf\implies{\frac{hf}{2}\:=\:hf_o - hf}$

$\sf\implies{\frac{hf}{2} + hf \:= \: hf_o}$

$\sf\implies{\frac{hf + 2hf }{2}\:=\:hf_o}$

$\sf\implies{\frac{3hf}{2}\:=\:hf_o}$

$\sf\implies{f_o\:=\: \frac{3f}{2}}$

Hence,The threshold frequency for the photoelectric emission $\sf{\frac{3f}{2}}$