Photoelectrons are emitted from a metal surface with velocity 2 . 0 × 10 5 ms - 1 . If a ray of 6000 A o be incident on the surface, then calculate the work function.

Given: Velocity of photoelectrons v = 2 . 0 × 10 5 m / s , Wave length of the incident radiations λ = 6000 × 10 - 10 m According to the equation of photoelectric emission the maximum energy of the emitted photoelectrons is given by E max = 1 2 mv 2 = hν - ϕ = hc λ = ϕ where m = mass of electron, hν = energy of incident radiations and ϕ = work function of the metal c , λ are the velocity and wavelength of light. Substituting the given values, we have 1 2 × 9 . 1 × 10 - 31 × 2 × 10 5 2 = 6 . 62 × 10 - 34 × 3 × 10 8 6000 × 10 - 10 - ϕ 0r ϕ = 3 . 118 × 10 - 19 1 . 6 × 10 - 19 = 1 . 95 eV Hence the work function is 1 . 95 eV

Photoelectrons are emitted from a metal surface with velocity 2 . 0 × 10 5 ms - 1 . If a ray of 6000 A o be incident on the surface, then calculate the work function.

Given: Velocity of photoelectrons v = 2 . 0 × 10 5 m / s , Wave length of the incident radiations λ = 6000 × 10 - 10 m According to the equation of photoelectric emission the maximum energy of the emitted photoelectrons is given by E max = 1 2 mv 2 = hν - ϕ = hc λ = ϕ where m = mass of electron, hν = energy of incident radiations and ϕ = work function of the metal c , λ are the velocity and wavelength of light. Substituting the given values, we have 1 2 × 9 . 1 × 10 - 31 × 2 × 10 5 2 = 6 . 62 × 10 - 34 × 3 × 10 8 6000 × 10 - 10 - ϕ 0r ϕ = 3 . 118 × 10 - 19 1 . 6 × 10 - 19 = 1 . 95 eV Hence the work function is 1 . 95 eV

When the light of 3000 A o is incident on sodium cathode, the stopping potential is 1 . 85 V ; and when the light of 4000 A o is incident, then the stopping potential becomes 0 . 82 V . Calculate Planck's constant, work function of sodium, threshold wavelength for sodium.

The photoelectric equation can be written as: hν = K max + ϕ where h , ν are the Planck's constant and the frequency of incident radiations and ϕ = work function of the metal. This can also be written as for two wave lengths as: hc λ 1 = ϕ + eV 1 and hc λ 2 = ϕ + eV 2 where V 1 . V 2 are the stopping potential for wavelength λ 1 a n d λ 2 we get by subtracting h c 1 λ 2 - 1 λ 1 = e V 2 - V 1 Given here, c = 3 × 10 8 m / s , λ 1 = 3 × 10 - 7 m and λ 2 = 4 × 10 - 7 m V 1 = 0 , 82 V V 2 = 1 . 85 V Substituting the values, h × 3 . 0 × 10 8 1 3 × 10 - 7 - 1 4 × 10 - 7 = 1 . 6 × 10 - 19 ( 1 . 85 - 0 . 82 ) or h × 3 . 0 × 10 × 1 12 × 10 - 7 = 1 . 6 × 10 - 19 ( 1 . 03 ) or h = 1 . 6 × 10 - 19 × 1 . 03 × 12 × 10 - 7 3 × 10 8 = 6 . 6 × 10 - 34 Js The maximum kinetic energy in the first case is given by: K max = hc λ - ϕ = 6 . 62 × 10 - 34 × 3 . 0 × 10 8 3 × 10 - 7 × 1 . 6 × 10 - 19 - 1 . 85 = 4 . 13 - 1 . 85 = 2 . 28 eV The stopping potential ( e V 0 ) will completely stop the electron with this energy so we have eV 0 = 2 . 28 eV or v 0 = 2 . 28 V The Stopping potential will be the light of wavelength corresponding to this potential i.e hc λ 0 = 2 . 28 × e or λ 0 = hc e × 2 . 28 Substituting the values, we have λ 0 = 6 . 62 × 10 - 34 × 3 × 10 8 2 . 28 × 1 . 6 × 10 - 19 = 5440 × 10 - 10 m

At stopping potential, if the wavelength of the incident light is kept fixed at 4000 A o , but the intensity of light is increased two times, will photoelectric current be obtained? Give reasons for your answer.

At stopping potential the photoelectrons are emitted but without any Kinetic energy, hence no current is produced. By increasing the intensity of the incident radiations, only the number of photons will increase without any change in their energy. Hence, no current will be obtained.