Question

this energy diagram shows the allowed energy levels of an electron in a certain atom. (note: the si prefix zepto means $10^{-21}$. you can find the meaning of any si prefix in the aleks data tab.)

use this diagram to complete the table below.

if the electron makes the transition shown by the red arrow, from c to a, calculate the wavelength of the photon that would be absorbed or emitted. round your answer to 3 significant digits.

Explanation:

Step1: Calculate the energy change

From the diagram, the energy of level \(C\) is \(500\space zJ\) and the energy of level \(A\) is \(200\space zJ\). The energy change \(\Delta E=E_{C}-E_{A}\).
\(\Delta E=(500 - 200)\space zJ=300\space zJ = 300\times10^{-21}\space J\)

Step2: Use the formula \(\Delta E=\frac{hc}{\lambda}\) to find the wavelength \(\lambda\)

We know that \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant) and \(c=3\times 10^{8}\space m/s\) (speed of light).
Rearrange the formula \(\lambda=\frac{hc}{\Delta E}\)
Substitute the values:
\(\lambda=\frac{6.626\times 10^{-34}\space J\cdot s\times3\times 10^{8}\space m/s}{300\times 10^{-21}\space J}\)
First, calculate the numerator: \(6.626\times10^{-34}\times3\times10^{8}=1.9878\times 10^{-25}\space J\cdot m\)
Then, \(\lambda=\frac{1.9878\times 10^{-25}\space J\cdot m}{300\times 10^{-21}\space J}\)
\(\lambda = 6.626\times10^{-7}\space m\)

Step3: Convert meters to nanometers

Since \(1\space m = 10^{9}\space nm\), then \(\lambda=6.626\times 10^{-7}\space m\times\frac{10^{9}\space nm}{1\space m}\)
\(\lambda = 663\space nm\)

Answer:

\(663\space nm\)