When light of wavelength $$300\,nm$$ (nanometre) falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, light of $$600\,nm$$ wavelength is sufficient for creating photoemission, what is the ratio of the work functions of the two emitters ?
A.
$$1:2$$
B.
$$2:1$$
C.
$$4:1$$
D.
$$1:4$$
Answer :
$$2:1$$
Solution :
Work function of a metal is $${W_0} = h{\nu _0}$$
$${\text{or}}\,{W_0} = \frac{{hc}}{{{\lambda _0}}}$$
Considering the two situation work function is given by
$$\therefore \frac{{{{\left( {{W_0}} \right)}_1}}}{{{{\left( {{W_0}} \right)}_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{600}}{{300}} = 2$$
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