Two photons of energy 4eV and 4.5 eV incident on two metals A and B respectively. Maximum kinetic energy for ejected electron is `T_A` for A and `T_B = T_A – 1.5 eV` for metal B. Relation between de-Broglie wavelength of ejected electron of A and B are `lambda_B = 2lambda_A` . The work function of metal B is–

Two photons of energy 4 eV and 4.5 eV are incident on two metals A and B respectively.The maximum kinetic energy for an ejected electron is T_(A) for A and T_(B)=T_(A)-1.5 eV for the metal B.The relation between the de-Broglie wavelengths of the ejected electron of A and B are lambda_(B)=2lambda_(B) .The work function of the metal B is

Photons of energies 4.25eV and 4.7eV are incident on two metal surfaces A and B respectively.The maximum KE of emitted electrons are respectively T_(A)eV and T_(B)=(T_(A)-1.5)eV .The ratio de-Broglie wavelengths of photoelectrons from them is lambda_(A):lambda_(B)=1.2 ,then find the work function of A and B

Photons of energies 4.25eV and 4.7eV are incident on two metal surfaces A and B respectively. The maximum KE of emitted electrons are respectively T_(A)eV and T_(B)=(T_(A)-1.5)eV . The ratio de-Broglie wavelengths of photo electrons from them is lamda_(A):lamda_(B)=1:2 , then find the work function of A and B

When photones of energy 4.0 eV fall on the surface of a metal A, the ejected photoelectrons have maximum kinetic energy T_(A) ( in eV) and a de-Broglie wavelength lambda_(A) . When the same photons fall on the surface of another metal B, the maximum kinetic energy of ejected photoelectrons is T_(B) = T_(A) -1.5eV . If the de-Broglie wavelength of these photoelectrons is lambda_(B) =2 lambda _(A) , then the work function of metal B is

Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be

Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be

Photons of energy 6 eV are incident on a metal surface whose work function is 4 eV . The minimum kinetic energy of the emitted photo - electrons will be

Photons of energy 1.5 eV and 2.5 eV are incident on a metal surface of work function 0.5 eV. What is the ratio of the maximum kinetic energy of the photoelectrons ?

photons of energy 4.25 eV strike the surface of metal A, the ejection photoelectric have maximum kinetic energy T_(A) eV energy 4.70 eV is T_(B) = (T_(A) - 1.50) eV if the de Brogle wavelength of these photoelec tron is lambda_(B) = 2 lambda_(A) , then