Photons

Question

1- A photon of frequency 8.2 x 10¹⁵ Hz is incident upon a photoelectric apparatus containing a metal whose threshold frequency is 3.6 x 10¹⁵ Hz. The stopping voltage is ___V.

2- The stopping voltage for an electron that has 7.30 x 10^-19 J of kinetic energy is ___V

Answer 1

(1)

The frequency of photon f = 8.2 x 10¹⁵ Hz .

The threshold frequency of metal is f₀ = 3.6 x 10¹⁵ Hz .

The stopping voltage can be evaluated using the formula E_k = q * V_stop .

V_stop  = E_k / q .

Where E_k is kinetic energy .

           q is the charge of electron ( 1.6 × 10–19 C )

The kinetic energy is E_k = E_p - w ⇒ hf - hf₀ ⇒ h( f -f₀ )

E_k = h( f -f₀ ) 

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

E_k = 6.63 × 10^-34 ( 8.2 - 3.6 )x 10¹⁵ .

E_k = 6.63 × 4.6 × 10^-19 ⇒ 30.498 × 10^-19  eV.

V_stop  = E_k / q ⇒ (30.498 × 10^-19 ) / (1.6 × 10–19 )

         = 19.06 V .

Hence , the  stopping voltage V_stop  = 19.06 V .

 

Answer 2

The kinetic energy  E_k = 7.30 x 10^-19 J .

The stopping voltage can be evaluated using the formula V_stop  = E_k / q .

Where E_k is kinetic energy .

           q is the charge of electron ( 1.6 × 10–19 C )

V_stop  = E_k / q  = (7.30 x 10^-19) / (1.6 × 10–19 C) = 4.562 V .

Hence , the  stopping voltage V_stop  = 4.562 V .