Question
The energy difference between the ground state of an atom and its excited state is $$3 \times {10^{ - 19}}\,J.$$ What is the wavelength of the photon required for this transition ?
A.
$$6.6 \times {10^{ - 34}}\,m$$
B.
$$3 \times {10^{ - 8}}\,m$$
C.
$$1.8 \times {10^{ - 7}}\,m$$
D.
$$6.6 \times {10^{ - 7}}\,m$$
Answer :
$$6.6 \times {10^{ - 7}}\,m$$
Solution :
$$\eqalign{
& \Delta E = \frac{{hc}}{\lambda } \cr
& \lambda = \frac{{6.6 \times {{10}^{ - 34}}Js \times 3 \times {{10}^8}\,m\,{s^{ - 1}}}}{{3 \times {{10}^{ - 19}}\,J}} \cr
& \,\,\,\,\, = 6.6 \times {10^{ - 7}}\,m \cr} $$