Part C An electron undergoes a transition from an initial (ni) to a final (nf )
ID: 906671 • Letter: P
Question
Part C An electron undergoes a transition from an initial (ni) to a final (nf ) energy state. The energies of the ni and nf energy states are 2.420×1019 J and 8.720×1020 J, respectively. Calculate the wavelength () of the light in nanometers (nm) corresponding to the energy change (E) value of this transition. You can use the following values for your calculations: Plancks constant (h)speed of light (c)1 m===6.626×1034 Js2.998×108 m/s. 1m=10^9nm. Express your answer in nanometers to four significant figures.
Explanation / Answer
Given
ni = 2.420×1019 J
nf = 8.720×1020 J,
Calculation of delta E
Delta E = E(final) – E(initial)
= 8.720×1020 J – (2.420×1019 J)
= 1.55 E-19 J
Delta E = hc/
Where h is planks constant , c is speed of light
We have to find (wavelength)
Lets rearrange above equation
= hc / delta E
= 6.626 E-34 Js x 2.998E8 m per s / 1.55 E-19 J
= 1.28 E-6 m
= 1.28 E-6 m
Lets find it in nm
= 1.28 E-6 x 1E9 nm / 1 m
= 1282.45 nm
in nm = 1282.45 nm