A hydrogen atom in an excited state emits a Balmer beta photon followed by a Lyman alpha photon. The electron decay from: the first excited state to the ground state, the second excited state to the ground state, the third excited state to the first excited state to the ground state, the third excited state to the second excited state to the ground state, or the third excited state to the second excited state to the first excited state to the ground state.
To determine the electron decay from different excited states in the hydrogen atom, let's review the basic principles behind the emission of photons.
1. Balmer series:
The Balmer series corresponds to electron transitions in the hydrogen atom where the electron starts in a higher energy level and drops to the second energy level (n = 2).
The Balmer series consists of several spectral lines, including the Balmer beta (Hβ) line. This line corresponds to the electron transitioning from the third energy level (n = 3) to the second energy level (n = 2).
2. Lyman series:
The Lyman series corresponds to electron transitions in the hydrogen atom where the electron starts in a higher energy level and drops to the first energy level (n = 1).
The Lyman series also consists of several spectral lines, including the Lyman alpha (Hα) line. This line corresponds to the electron transitioning from the third energy level (n = 3) to the first energy level (n = 1).
Now, let's analyze the given scenarios:
a) The first excited state to the ground state:
If the electron transitions from the first excited state (n = 2) to the ground state (n = 1), it would emit a Lyman alpha photon (Hα). However, this scenario is not mentioned in the given options.
b) The second excited state to the ground state:
If the electron transitions from the second excited state (n = 3) to the ground state (n = 1), it would emit a Lyman alpha photon (Hα).
c) The third excited state to the first excited state to the ground state:
This scenario involves the electron transitioning from the third excited state (n = 4) to the first excited state (n = 2) and then to the ground state (n = 1). The emitted photons would be a Balmer beta photon (Hβ) followed by a Lyman alpha photon (Hα).
d) The third excited state to the second excited state to the ground state:
This scenario involves the electron transitioning from the third excited state (n = 4) to the second excited state (n = 3) and then to the ground state (n = 1). The emitted photons would be a Balmer beta photon (Hβ) followed by a Lyman alpha photon (Hα). This scenario matches the given sequence of emitted photons.
e) The third excited state to the second excited state to the first excited state to the ground state:
This scenario involves the electron transitioning from the third excited state (n = 4) to the second excited state (n = 3), then to the first excited state (n = 2), and finally to the ground state (n = 1). The emitted photons would be a Balmer beta photon (Hβ) followed by a Lyman alpha photon (Hα).
Therefore, the correct answer to the given options is: The third excited state to the second excited state to the ground state.