An electron falls from an energy level of -5.44eV to another energy level, E. If the emitted photon is of…

1. Analyze the Question

We are given an electron transitioning between two energy levels. We need to find the energy change ($\Delta E$) of the emitted photon, given its wavelength ($\lambda$), Planck’s constant ($h$), and the speed of light ($c$).

2. Explain the Concept

When an electron moves from a higher energy level to a lower one, it emits a photon. The energy of this photon is exactly equal to the difference in energy between the two levels. The relationship between energy and wavelength is given by the formula:

Where:

• $E$ = Energy in Joules (J)
• $h$ = Planck’s constant ($6.63 \times 10^{-34} \text{ Js}$)
• $c$ = Speed of light ($3.0 \times 10^8 \text{ ms}^{-1}$)
• $\lambda$ = Wavelength ($5.68 \times 10^{-6} \text{ m}$)

3. Show the Solution

Step 1: Plug the values into the formula.

Step 2: Multiply the numerator.

Step 3: Divide by the wavelength.

Comparing this result to the provided options, $3.50 \times 10^{-20} \text{ J}$ is closest to $3.49 \times 10^{-20} \text{ J}$.

4. Identify the Answer

The calculated energy change is approximately $3.49 \times 10^{-20} \text{ J}$. This matches option C.

5. Exam Tip

When solving physics problems involving very small or large numbers, always handle the powers of 10 separately from the coefficients. This prevents calculation errors. Also, remember that $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$; if the question had asked for the answer in electron-volts, you would divide your final answer by $1.6 \times 10^{-19}$.