Question

q7. one of the emissions by sr atoms in red fireworks (and emergency flares) has a frequency of 4.62×10^14 hertz. what is the difference in quantum energy levels in joules for the transition that is responsible for this prominent red colored line in the strontium spectrum?
a) 4.62×10^19 j.
b) 3.60×10^19 j.
c) 3.06×10^19 j.
d) 6.63×10^19 j.

Explanation:

Step1: Recall the Planck - Einstein relation

The energy of a photon is given by $E = h
u$, where $h$ is Planck's constant ($h=6.63\times 10^{-34}\text{ J}\cdot\text{s}$) and $
u$ is the frequency of the photon. The difference in quantum energy levels $\Delta E$ for a transition is equal to the energy of the emitted photon.

Step2: Substitute the values

We are given $
u = 4.62\times 10^{14}\text{ Hz}$. Substitute $h = 6.63\times 10^{-34}\text{ J}\cdot\text{s}$ and $
u=4.62\times 10^{14}\text{ Hz}$ into the formula $E = h
u$.
$\Delta E=(6.63\times 10^{-34}\text{ J}\cdot\text{s})\times(4.62\times 10^{14}\text{ Hz})$
$\Delta E = 6.63\times4.62\times10^{-34 + 14}\text{ J}$
$\Delta E=30.6306\times 10^{-20}\text{ J}$
$\Delta E = 3.06306\times 10^{-19}\text{ J}\approx3.06\times 10^{-19}\text{ J}$

Answer:

c) $3.06\times 10^{-19}\text{ J}$