A) 2.5 eV, 0
B) 4.9 eV, 3eV
C) 5.9 eV, 0.5 eV
D) 5.5 eV, 3eV
Correct Answer: B
Solution :
Energy of the incident light \[\frac{hc}{\lambda }=\frac{1240\,eV\,nm}{280\,nm}=4.4\,eV\] \[{{(KE)}_{\max }}=\frac{hc}{\lambda }-{{\phi }_{0}}=4.4-2.5=1.9\,eV.\] So, maximum K.E. of the ejected photoelectrons is 1.9 eV KE acquired by these electrons when accelerated under a potential difference of 3V is = 3 eV. \[\therefore \]\[{{(KE)}_{max}}\]of the photoelectrons = 1.9 + 3 = 4.9 eV The electrons which are just ejected from the cathode will have minimum KE at the anode, so, minimum kinetic energy of the photoelectrons must be 3 eV Hence, the correction option is (b).
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