Answer :
Final answer:
The Hydrogen atom absorbs photon of wavelength 97.3 nm, causing electron to move to 3rd energy level. When the electron relaxes back to 2nd energy level, the emitted photon would be in the visible range. The exact wavelength can be calculated using Rydberg's formula.
Explanation:
In this scenario the given wavelength of 97.3 nm corresponds to the Lyman series, exciting the electron from the ground state (n=1) to the third energy level (n=3), given the series pattern in the problem's information. The photon emitted as the electron relaxes from the third level back to ground state will be of longer wavelength and less energy, making it in the visible range based on the Bohr model.
Using the formula 1/λ = R(1/n1^2 - 1/n2^2) where R is Rydberg constant, n1 is the lower energy level and n2 is the higher, we can calculate the wavelength of the emitted photon by substituting the known values. In this scenario, n1=1 (ground state) and n2=2 (first excited state as it is in visible range).
So, by substituting the values to get the required wavelength in the Balmer series (visible range), the electron would need to relax back to the second energy level (n=2). The final transition from second level to ground state would then produce a photon in the visible light spectrum.
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Final answer:
The wavelength of the emitted photon cannot be determined without knowing the final energy level to which the electron jumps. However, it will be greater than 97.3 nm.
Explanation:
To find the wavelength of the emitted photon, we need to determine the energy levels involved. In this case, the hydrogen atom is initially in the ground state, which corresponds to n₁ = 1. The unknown energy level to which the electron jumps is not specified.
Using the Rydberg formula, we can calculate the energy difference between the initial and final energy levels:
1/λ = R * (1/n₁² - 1/n₂²)
Since n₁ = 1, the equation simplifies to:
1/λ = R * (1 - 1/n₂²)
To find the wavelength of the emitted photon, we need to determine the value of n₂. Unfortunately, the question does not provide this information. Therefore, we cannot calculate the exact wavelength of the emitted photon without knowing the final energy level to which the electron jumps.
However, we can make some general observations. When an electron jumps from a higher energy level to a lower energy level, the emitted photon has a longer wavelength than the absorbed photon. This means that the wavelength of the emitted photon will be greater than 97.3 nm.
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