Light of wavelength $$200\,\mathop {\text{A}}\limits^ \circ $$ fall on aluminium surface. Work function of aluminium is $$4.2\,eV.$$ What is the kinetic energy of the fastest emitted photoelectrons?
A.
$$2\,eV$$
B.
$$1\,eV$$
C.
$$4\,eV$$
D.
$$0.2\,eV$$
Answer :
$$2\,eV$$
Solution :
By Einstein’s equation of photo-electric effect, the maximum kinetic energy of emitted photo-electrons is given by
$${E_k} = hv - W\,\,{\text{or}}\,\,{E_k} = \frac{{hc}}{\lambda } - W$$
Where, $$h$$ = Planck’s constant
$$v$$ = frequency of incident light
$$W$$=work function of metal
$$\lambda $$ = wavelength of incident light
$$\eqalign{
& {E_k} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2000 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}}eV - 4.2\,eV \cr
& {\text{So,}}\,{E_k} = 2\,eV \cr} $$
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