Question

An electron of an atom jumps down from one energy level to another and emits a photon that contains 8.83 x 10^-18joule of energy. What was the frequency of the emitted electromagnetic energy?(a) 1.33 x 10^16 Hz(b) 6.74 x 10^15 Hz(c) 3.27 x 10^-15 Hz(d) 7.5 x 10^-17 Hz

Answer 1

In order to determine the frequency of the emmited electromagnetic energy, use the following formula:

`E=hf`

where:

E: energy of the photon = 8.83*10^(-18) J

h: Planck's constant = 6.63*10^(-34)J*s

f: frequency of the photon = ?

solve the previous equation for f, and replace the values of the other parameters, as follow:

`f=(E)/(h)=(8.83\cdot10^(-18)J)/(6.63\cdot10^(-34)J\cdot s)=1.33\cdot10^(16)Hz`