Question
When electronic transition occurs from higher energy state to lower energy state with energy difference equal to $$\Delta E$$ electron volts, the wavelength of the line emitted is approximately equal to
A.
$$\frac{{12395}}{{\Delta E}} \times {10^{ - 10}}m$$
B.
$$\frac{{12395}}{{\Delta E}} \times {10^{10}}m$$
C.
$$\frac{{12395}}{{\Delta E}} \times {10^{ - 10}}cm$$
D.
$$\frac{{12395}}{{\Delta E}} \times {10^{10}}cm$$
Answer :
$$\frac{{12395}}{{\Delta E}} \times {10^{ - 10}}m$$
Solution :
$$\eqalign{
& \lambda = \frac{{hc}}{{\Delta E}} \cr
& = \frac{{6.62 \times {{10}^{ - 34}}J\,s \times 3 \times {{10}^8}m{s^{ - 1}}}}{{E \times 1.602 \times {{10}^{ - 19}}J}} \cr
& = \frac{{12395}}{E} \times {10^{ - 10}}m \cr} $$