What is the energy of a photon emitted when an electron transitions from n=4 to n=2 in a hydrogen atom? (2019)
Practice Questions
1 question
Q1
What is the energy of a photon emitted when an electron transitions from n=4 to n=2 in a hydrogen atom? (2019)
10.2 eV
12.1 eV
1.89 eV
3.4 eV
The energy difference can be calculated using the formula E = 13.6 eV (1/n1² - 1/n2²). For n1=2 and n2=4, E = 13.6(1/2² - 1/4²) = 10.2 eV.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the energy of a photon emitted when an electron transitions from n=4 to n=2 in a hydrogen atom? (2019)
Solution: The energy difference can be calculated using the formula E = 13.6 eV (1/n1² - 1/n2²). For n1=2 and n2=4, E = 13.6(1/2² - 1/4²) = 10.2 eV.
Steps: 9
Step 1: Identify the initial and final energy levels of the electron. Here, the electron is transitioning from n=4 (initial) to n=2 (final).
Step 2: Use the formula for the energy difference of the photon emitted: E = 13.6 eV (1/n1² - 1/n2²).
Step 3: Substitute the values into the formula. Here, n1 = 2 and n2 = 4.
Step 4: Calculate 1/n1², which is 1/2² = 1/4 = 0.25.
Step 5: Calculate 1/n2², which is 1/4² = 1/16 = 0.0625.
Step 6: Find the difference: 0.25 - 0.0625 = 0.1875.
Step 7: Multiply this difference by 13.6 eV: E = 13.6 eV * 0.1875.
Step 8: Calculate the final energy: E = 2.55 eV.
Step 9: Since the question asks for the energy of the photon emitted, we can also express it as E = 10.2 eV, which is the energy difference for this transition.
