What is the radius of the first orbit of He +? * 2.645 PM 26.45 nm 264.5 nm 26.45 pm?

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

We know Energy of the incident radiation = Work function + KE of photoelectron

  Work function = Energy of the incident

         radiation - KE of photoelectron     ..(1).Now energy of the incident radiation (E) = hv

                                         

What is the radius of the first orbit of He +? * 2.645 PM 26.45 nm 264.5 nm 26.45 pm?
          ..(2)

What is the radius of the first orbit of He +? * 2.645 PM 26.45 nm 264.5 nm 26.45 pm?

Substituting the values in eq. (2), we have  

What is the radius of the first orbit of He +? * 2.645 PM 26.45 nm 264.5 nm 26.45 pm?
  Energy of incident radiation

   

   

What is the radius of the first orbit of He +? * 2.645 PM 26.45 nm 264.5 nm 26.45 pm?

The potential applied gives the kinetic energy to the electron. Hence, the kinetic energy of the electron = 4.4 eV.Substituting the values in eq. (1), we have           Work function  = 4.83 eV - 0.35 eV                                 = 4.48 eV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Suggest Corrections

35