When photons of energy 4.25 eV strike the surface of a metal A , the ejected photoelectrons have maximum kinetic energy, T _ A expressed in 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-1.50 eV ) . If the de Broglie wavelength of these photoelectrons is _B=2 _A , then :

Step-by-step solution

aligned & h v= K.E. (T)+ work function ( W ) & hv = T + W & 4.25 eV = T _ A + W _ A ( for Metal A) & 4.70 eV = T _ B + W _ B ( for Metal B) aligned Since T_B= (T_A-1.5 ) eV Also = h / p aligned & = h 2 mT p ^2 2 ~m = T = K.E. & _ A _ B = T _ B T _ A aligned Since _A= 1 2 _B aligned & T_A=4 T_B & T_B=T_A-1.50 gives & T_B=4 T_B-1.5 & T_B=0.5 eV T_A & =2 eV W_A & =2.25 eV & W_B=4.20 EV aligned