Einstein’s work on photoelectric effect gives support to
A.
$$E = m{c^2}$$
B.
$$E = h\nu $$
C.
$$h\nu = \frac{1}{2}m{\nu ^2}$$
D.
$$E = \frac{h}{\lambda }$$
Answer :
$$E = h\nu $$
Solution :
In 1905, Einstein realized that the photoelectric effect could be understood if the energy in light is not spread out over wavefronts but is concentrated in small packets, or photons. Each photon of light of frequency $$\nu $$ has the energy $$h\nu .$$ Thus, Einstein's work on photoelectric effect gives support to $$E = h\nu .$$
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