An electron and proton are separated by a large distance. The electron starts approaching the proton with energy 3 eV. The proton captures the electron and forms a hydrogen atom in second excited state. The resulting photon is incident on a photosensitive metal of threshold wavelength 4000 Å. What is the maximum kinetic energy of the emitted photoelectron?
Show Hint
Remember the energy level conventions: "second excited state" means the principal quantum number is n=3. Also, the formula $E(\text{eV}) \approx 1240/\lambda(\text{nm})$ is a very useful shortcut for photoelectric effect and atomic physics problems.
To find the maximum kinetic energy of the emitted photoelectron, we need to follow these steps:
The electron approaches the proton and forms a hydrogen atom in a specific excited state. Given that the electron has an initial energy of 3 eV, and it is said to form a hydrogen atom in the second excited state, we need to understand what this means.
In a hydrogen atom, the second excited state corresponds to the n=3 level. The energy levels in hydrogen are given by the formula:
E_n = - \frac{13.6}{n^2} \text{ eV}
where n represents the principal quantum number.
For n=3, the energy level is:
E_3 = - \frac{13.6}{3^2} = -1.51 \text{ eV}
Since the electron started with 3 eV, it must release additional energy as a photon when transitioning to this energy level.
The initial energy of the electron is 3 eV, so the energy of the emitted photon when the electron transitions from free to n=3 is:
E_{\text{photon}} = 3 - (-1.51) = 3 + 1.51 = 4.51 \text{ eV}
This photon is then incident on a photosensitive metal with a threshold wavelength of 4000 Å. This threshold wavelength corresponds to the minimum energy required to eject photoelectrons, calculated using:
\lambda_{\text{threshold}} = 4000 \Å = 4000 \times 10^{-10} \text{ m}
The energy corresponding to this wavelength (E_{\text{threshold}}) is given by:
E_{\text{threshold}} = \frac{hc}{\lambda_{\text{threshold}}}
Where the Planck's constant (h) is 6.63 \times 10^{-34} \text{ J s} and speed of light (c) is 3 \times 10^8 \text{ m/s}.
Convert the energy to eV: E_{\text{threshold}} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} = 3.1 \text{ eV}
The maximum kinetic energy (KE_{\text{max}}) of the emitted photoelectron is the energy of the photon minus the threshold energy:
KE_{\text{max}} = E_{\text{photon}} - E_{\text{threshold}} = 4.51 - 3.1 = 1.41 \text{ eV}
Therefore, the maximum kinetic energy of the emitted photoelectron is 1.41 eV, matching the given correct answer.
