The cathode of a photoelectric cell is changed such that the work function changes from $${W_1}$$ to $${W_2}\left( {{W_2} > {W_1}} \right).$$
If the current before and after changes are $${I_1}$$ and $${I_2},$$ all other conditions remaining unchanged, then (assuming $$h\nu > {W_2}$$ )
A.
$${I_1} = {I_2}$$
B.
$${I_1} < {I_2}$$
C.
$${I_1} > {I_2}$$
D.
$${I_1} < {I_2} < 2{I_1}$$
Answer :
$${I_1} = {I_2}$$
Solution :
By work function of a metal, it means that the minimum energy required for the electron in the highest level of conduction band to get out of the
metal. The work function has no effect on photoelectric current as long as $$h\nu > {W_0}.$$ The photoelectric current is proportional to the intensity of incident light. Since, there is no change in the intensity of light, hence $${I_1} = {I_2}$$
Releted MCQ Question on Modern Physics >> Dual Nature of Matter and Radiation
Releted Question 1
A particle of mass $$M$$ at rest decays into two particles of
masses $${m_1}$$ and $${m_2},$$ having non-zero velocities. The ratio of the de Broglie wavelengths of the particles, $$\frac{{{\lambda _1}}}{{{\lambda _2}}},$$ is
A proton has kinetic energy $$E = 100\,keV$$ which is equal to that of a photon. The wavelength of photon is $${\lambda _2}$$ and that of proton is $${\lambda _1}.$$ The ration of $$\frac{{{\lambda _2}}}{{{\lambda _1}}}$$ is proportional to