Question
The energy of a photon of light is $$3\,eV.$$ Then the wavelength of photon must be
A.
$$4125\,nm$$
B.
$$412.5\,nm$$
C.
$$41250\,nm$$
D.
$$4\,nm$$
Answer :
$$412.5\,nm$$
Solution :
If energy $$E$$ is expressed in $$\left( {eV} \right)$$ and wavelength $$\lambda $$ (in $$\mathop {\text{A}}\limits^ \circ $$ ), then energy of photon,
$$\eqalign{
& E = \frac{{hC}}{\lambda } = \frac{{12375}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}eV \cr
& \therefore \lambda = \frac{{12375}}{{E\left( {eV} \right)}}\mathop {\text{A}}\limits^ \circ \,\,\left[ {\because hc = 12375\,eV - \mathop {\text{A}}\limits^ \circ } \right] \cr
& = \frac{{12375}}{{3\left( {eV} \right)}}\mathop {\text{A}}\limits^ \circ = 4125\,\mathop {\text{A}}\limits^ \circ \cr
& = 412.5\,nm \cr} $$
NOTE
Energy of photon is $$E = \frac{{hc}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}} = \frac{{12375}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}eV$$
Here, $$hc = 12375\,eV - \mathop {\text{A}}\limits^ \circ $$ comes from the following procedure
$$\eqalign{
& hc = \left( {{\text{Planck's}}\,{\text{constant}}} \right)\left( {{\text{velocity}}\,{\text{of}}\,{\text{light}}} \right) \cr
& = \frac{{\left( {6.6 \times {{10}^{ - 34}}\;J - s} \right)\left( {3 \times {{10}^8}\;m/s} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\,J/eV} \right)}} \cr
& = 12.375 \times {10^{ - 7}}eV - m \cr
& = 12375\,eV - \mathop {\text{A}}\limits^ \circ \cr} $$