Question

question #3
the energy level diagram below shows the allowed energy levels of an electron in an atom of element x.
calculate the wavelength of a photon emitted by the transition from c to b. m

Explanation:

Step1: Calculate the energy difference

The energy of the photon emitted is the difference in energy between level C and level B.
$$\Delta E=E_{C}-E_{B}$$
Substitute $E_{C}=- 1.1\times10^{-20}\text{ J}$ and $E_{B}=-2.3\times10^{-20}\text{ J}$
$$\Delta E=(-1.1\times 10^{-20})-(-2.3\times 10^{-20}) = 1.2\times10^{-20}\text{ J}$$

Step2: Use the formula $\Delta E = h

u=\frac{hc}{\lambda}$
We know that $h = 6.63\times10^{-34}\text{ Js}$ and $c=3\times10^{8}\text{ m/s}$.
From $\Delta E=\frac{hc}{\lambda}$, we can solve for $\lambda$:
$$\lambda=\frac{hc}{\Delta E}$$
Substitute the values:
$$\lambda=\frac{6.63\times 10^{-34}\times3\times 10^{8}}{1.2\times 10^{-20}}$$
First, calculate the numerator: $6.63\times10^{-34}\times3\times10^{8}=1.989\times10^{-25}$
Then, $\lambda=\frac{1.989\times 10^{-25}}{1.2\times 10^{-20}}=1.6575\times10^{-5}\text{ m}$

Answer:

$1.66\times 10^{-5}\text{ m}$