Question

The work function of Cesium is $2.27eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600nm$ wavelength is

• A. $0.5V$
• B. $-0.2V$
• C. $-0.5V$
• D. $0.2V$
• E. None of the above

Answer 1

Answer: (E)

None of the above

Energy of photon $=\frac{hc}{\lambda }=2.066eV$

which is less than the work function of Cesium. Hence the concept of cut off voltage has no meaning here.