Photon of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV . The stopping voltage required for these electrons are

Photons of 5.5 eV energy fall on the surface of the metal emitting photoelectrons of maximum kinetic energy 4.0 eV. The stopping voltage required for these electrons is

When a radiation of energy 5eV falls on a surface,the emitted photoelectrons have a maximum kinetic energy of 3eV .The stopping potential is

Photons of energy 7eV fall on the surface of a metal X resulting in emission of photoelectrons having maximum kinetic energy E_(1) = 1eV . Y is another metal on the surface of which photons of energy 6eV are incident and result in emission of photoelectrons of maximum kinetic energy E_(2)= 2eV . The ratio of work functions of metals X and Y, (phi_(x))/(phi_(y)) is ____________

Photons of energy 7eV fall on the surface of a metal X resulting in emission of photoelectrons having maximum kinetic energy E_(1) = 1eV . Y is another metal on the surface of which photons of energy 6eV are incident and result in emission of photoelectrons of maximum kinetic energy E_(2)= 2eV . The ratio of work functions of metals X and Y, (phi_(x))/(phi_(y)) is ____________

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

When photon of energy 25eV strike the surface of a metal A, the ejected photelectron have the maximum kinetic energy photoelectrons have the maximum kinetic energy T_(A)eV and de Brogle wavelength lambda_(A) .The another kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.76 eV is T_(B) = (T_(A) = 1.50) eV .If the de broglie wavelength of these photoelectrons is lambda_(B) = 2 lambda_(A) then i. (W_(B))_(A) = 2.25 eV II. (W_(0))_(B) = 4.2 eV III T_(A) = 2.0 eV IV. T_(B) = 3.5 eV