Photons with energy 5 eV are incident on a cathode C in a photoelectric cell. The maximum energy of emitted photoelectrons is 2 eV. When photons of energy 6 eV are incident on C, no photoelectrons will reach the anode A, if the stopping potential of A relative to C is
–3 V
+3 V
+4 V
–1 V
Explanation:
$k_{max} = h\nu – \phi$
$2eV = 5eV – \phi$
$\phi = 3eV$
So Vst = 3 volt
Vcathode – Vanode = 3 volt
Vanode – Vcathode = – 3 volt