photoelectric surface

Question

1- A photoelectric surface has a work function of 3.30x10^-19 J. The threshold frequency of this surface is ___x10¹⁴ Hz 

2- A photoelectric surface requires a light of maximum wavelength 675 nm to cause electron emission. The work function of this surface is ___eV

3-Light with a wavelength of 530 nm falls on a photoelectric surface that has a work function of 1.7 eV. The maximum kinetic energy of any emitted photoelectrons is ___eV

4-  A photoelectric surface has a work function of 2.75 eV. The minimum frequency of light that will cause photoelectron emission from this surface is x10¹⁴ Hz

Answer 1

(1)

Work function of a photoelectric surface w = 3.30x10^-19 J .

Work function w =  h * f₀ .

f₀  = w / h

Where h is plank's constant = 6.63 × 10^-34 J·sec .

         f₀ is  is the threshold frequency .

f₀  = ( 3.30x10^-19 ) / (6.63 × 10^-34 ) 

f₀  = 4.977 × 10¹⁴ Hz .

The threshold frequency is 4.977 × 10¹⁴ Hz .

Answer 2

(2)

The maximum wave length of the λ = 675 nm = 675 × 10^-9 m .

Work function of a photoelectric surface w = h * f₀ ⇒ (h * c) / λ .      [Since f₀ = c / λ ]

Where h is plank's constant = 6.63 × 10^-34 J·sec .

         c = velocity of light = 3 × 10⁸ m/s .

w = (6.63 × 10^-34  × 3 × 10⁸ )/ 675 × 10^-9

w = 2.9449 × 10^-19 J

w = 1.83806 eV .

work function of this surface is 1.83806 eV .

Answer 3

(3)

The wave length of photon λ=530 nm = 530 × 10^-9 m .

Work function of a photoelectric surface w = 1.7 eV  = 2.7237 x 10^-19 J .

The kinetic energy is E_k = E_p - w ⇒ hf - w ⇒ [ (h*c)/λ ] - w .

E_k = [ (h*c)/λ ] - w 

E_k = [(6.63 × 10^-34  × 3 × 10⁸ )/ 530 × 10^-9 ] - 2.7237 x 10^-19 J

E_k = 3.7506x 10^-19 - 2.7237 x 10^-19 

E_k = 1.0269 x 10^-19  J 

E_k = 0.64088 eV .

The maximum kinetic energy of any emitted photoelectrons is 0.64088 eV .

Answer 4

(4)

Work function of a photoelectric surface w =2.75 eV = 4.40598 x10^-19 J .

Work function w =  h * f₀ .

f₀  = w / h

Where h is plank's constant = 6.63 × 10^-34 J·sec .

         f₀ is  is the threshold frequency or minimum frequency  .

f₀  = ( 4.40598 x 10^-19 ) / (6.63 × 10^-34 ) 

f₀  = 6.6494 × 10¹⁴ Hz .

The minimum frequency is 6.6494 × 10¹⁴ Hz .