An electron falls from an energy level of -5.44eV to another energy level, E. If the emitted photon is of wavelength 5.68 x 10\(^{-6}\)m, calculate the energy change. [ Plank's constant = 6.63 x 10\(^{-34}\)Js, emitted radiation speed = 3.0 x 10\(^8\)ms\(^{-1}\)]
A
1.49 x 10\(^{-20}\)J
B
1.49 x 10\(^{-19}\)J
C
3.49 x 10\(^{-20}\)J
D
3.49 x 10\(^{-19}\)J
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
Energy of the emitted photon(E) = \(\frac{\text{hc}}{\lambda}\)
\(\lambda\) = 5.68 x 10\(^{-6}\)m
h = 6.63 x 10\(^{-34}\)Js
c = 3.0 x 10\(^8\)ms\(^{-1}\)
E =\(\frac{(6.63\times 10^{-34}) \times (3.0 x 10^8)}{5.68\times 10^{-6}}\) = \(\frac{1.989\times 10^{-25}}{5.68\times 10^{-6}}\)
E = 3.50 x 10\(^{-20}\)J
Therefore, the energy change = 3.50 x 10\(^{-20}\)J
Source: WAEC / JAMB / NECO Classroom Archive (Record #70480)