When photoelectrons of energy 4.25 eV strike the surface of metal A , the ejected photoelectrons have maximum kinetic energy T _ A eV and de-Broglie wavelength _ A . The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is T _ B = ( T _ A -1.5 ) eV . If de-Broglie wavelength of these photoelectrons is _ B =2 _ A , then

Step-by-step solution

aligned & p _ A = h _ A , ~T _ A = P _ A ^2 2 ~m = h ^2 2 ~m _ A ^2 , ~T _ B = h ^2 2 ~m _ B ^2 = h ^2 8 ~m _ A ^2 & ~h ^2 2 ~m _ A ^2 =4.25- _ A ...(i) aligned aligned & h ^2 8 ~m _ A ^2 =4.70- _ B ...(ii) & Also, h ^2 8 ~m _ A ^2 = h ^2 2 ~m _ A ^2 -1.5 & 3 8 ~h ^2 ~m _ A ^2 =1.5 ~h ^2 ~m _ A ^2 =4 eV & ~T _ A =2 eV , ~T _ V =0.5 eV & _ A =2.25 eV , _ B =4.70-0.5=4.20 eV aligned