When a certain metal was irradiated with a light of frequency 3.2 * 10 to d power 16 Hz, the photo electrons had twice the K.E as emitted when d same metal was irradiated with light of frequency 2*10 to d power 16 Hz. Calculate the threshold frequency of the metal.

we know that,

E(Total Energy)=E_o(Minimum Energy reqd. for the electron)+K.E(kinetic Energy of Photo electron)

hv=hv₀ + 1/2 mv²

6.6*10^-34*(3.2 * 10¹⁶) =E_o+ K.E

  2.112*10^-17 =E_o+ K.E......................(1)

6.6*10^-34*(2*10¹⁶⁾ =E_o+ 2*K.E.

1.32*10-¹⁷=Eo+ 2 K.E  .......................(2)

solving equation (1) and (2):

E_o=2.904*10^-17 j

 

hvo=2.904*10^-17

 

vo=4.4*10¹⁶ Hz.