Question
The energy of photon is given as :
$$\Delta e/atom = 3.03 \times {10^{ - 19}}J\,ato{m^{ - 1}},$$ then the wavelength $$\left( \lambda \right)$$ of the photon is
( Given, $$h$$ (Planck’s constant) $$ = 6.63 \times {10^{ - 34}}Js,$$ $$c$$ (velocity of light) $$ = 3.00 \times {10^8}m{s^{ - 1}}$$ )
A.
6.56$$\,nm$$
B.
65.6$$\,nm$$
C.
656$$\,nm$$
D.
0.656$$\,nm$$
Answer :
656$$\,nm$$
Solution :
$${\text{According to formula,}}$$ $$E = \frac{{hc}}{\lambda }\left( {v = \frac{c}{\lambda }} \right)$$
$$\eqalign{
& {\text{Energy}}\,E = h\nu \cr
& {\text{3}}{\text{.03}} \times {10^{ - 19}} = \frac{{hc}}{\lambda } \cr
& \lambda = \frac{{6.63 \times {{10}^{ - 34}} \times 3.0 \times {{10}^8}}}{{3.03 \times {{10}^{ - 19}}}} \cr
& \,\,\,\,\, = 6.56 \times {10^{ - 7}}m \cr
& \,\,\,\,\, = 6.56 \times {10^{ - 7}} \times {10^9}nm \cr
& \,\,\,\,\, = 6.56 \times {10^2}nm \cr
& \,\,\,\,\, = 656\,nm \cr} $$