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Answer :
Final Answer:
The wavelength of light emitted in the second relaxation from n = 6 to n = 1 is approximately 121.6 nm. Therefore, the correct answer is A) 121.6 nm.
Explanation:
To solve this problem, we can use the Rydberg formula, which relates the wavelength of light emitted by a hydrogen-like atom to the initial and final energy levels:
1/λ = R_H (1/n_f² - 1/n_i²)
Where:
- λ is the wavelength of the emitted light,
- R_H is the Rydberg constant for hydrogen (1.097 x 10⁷ m⁻¹),
- n_f is the final energy level, and
- n_i is the initial energy level.
Given that the initial energy level n_i = 6 and the final energy level n_f = 1, we can first use the given wavelength (273.6 nm) to find the initial energy level.
1/λ = R_H (1/n_f² - 1/n_i²)
1/λ = R_H (1/1² - 1/6²)
Now, solve for λ using the given wavelength:
λ = 1/R_H ( 1/1² - 1/6²)
λ = 1/1.097 x 10⁷ m⁻¹ ( 1 - 1/36)
λ ≅ 121.6 nm
So, the wavelength of light emitted in the second relaxation from n = 6 to n = 1 is approximately 121.6 nm.
Therefore, the correct answer is A) 121.6 nm.
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