A metal surface is illuminated by a light of given intensity and frequency to cause photo emission. If the intensity of illumination is reduced to one - fourth of its original value, then the maximum $$KE,$$ of emitted photoelectrons will become
A.
$${\left( {\frac{1}{{16}}} \right)^{th}}$$ of original value
B.
unchanged
C.
twice the original value
D.
four times the original value
Answer :
unchanged
Solution :
Maximum $$KE$$ depends on the frequency of incident radiation, not on intensity.
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