If the threshold wavelength for a certain metal is $$2000\,\mathop {\text{A}}\limits^ \circ ,$$ then the work function of the metal is
A.
$$6.2\,J$$
B.
$$6.2\,eV$$
C.
$$6.2\,MeV$$
D.
$$6.2\,keV$$
Answer :
$$6.2\,eV$$
Solution :
The minimum energy required to remove an electron from the surface of metal without giving any kinetic energy is called work function.
$${W_0} = h{\nu _0} = \frac{{hc}}{{{\lambda _0}}}$$
Given wavelength,
$$\eqalign{
& {\lambda _0} = 2000\,\mathop {\text{A}}\limits^ \circ \cr
& = 2000 \times {10^{ - 10}}m \cr
& \therefore {W_0} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2000 \times {{10}^{ - 10}}}} \cr
& = 9.9 \times {10^{ - 19}}J \cr
& = \frac{{9.9 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\,\left[ {1\,eV = 1.6 \times {{10}^{ - 19}}\,J} \right] \cr
& = 6.2\,eV \cr} $$
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