Electron exhibiting a particle wavelength of 4.02 x 10-9 are emitted from hydrogen atoms after interaction with high energy photons. Assuming that the electrons in the hydrogen atom were in their ground state prior to interaction, determine the wavelength of the photon which generated the free electrons
Determine the minimum potential that must be applied to a proton so that, on interaction with a hydrogen atom, it can excite the ground state electron to a state of n=2
could you tell me the above answer?
12.7 for Determine the minimum potential that must be applied to a proton so that, on interaction with a hydrogen atom, it can excite the ground state electron to a state of n=4
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To determine the wavelength of the photon that generated the free electrons, we can use the concept of the energy difference between the initial and final states of the electron in the hydrogen atom.
In this case, the electrons were in their ground state (n=1) before interaction and were emitted with a particle wavelength of 4.02 x 10^(-9) m. We can use the energy-wavelength relationship for particles, which is given by the formula:
λ = h / p
Where λ is the wavelength, h is the Planck's constant (6.626 x 10^(-34) J·s), and p is the momentum of the particle. Since we are dealing with particles, the momentum can also be expressed in terms of the De Broglie wavelength:
p = h / λ
Substituting the given wavelength (4.02 x 10^(-9) m) into the above equation, we can calculate the momentum of the electron:
p = (6.626 x 10^(-34) J·s) / (4.02 x 10^(-9) m)
p ≈ 1.646 x 10^(-25) kg·m/s
Now, let's consider the energy difference between the initial (ground) state and the final state of the electron. According to the energy level transition in hydrogen atoms, the energy difference (ΔE) between two energy levels can be given by:
ΔE = E_final - E_initial
ΔE = -13.6 eV - (-13.6 eV)
ΔE = 0 eV
Since the energy difference is zero, it means that the energy of the photon absorbed is equal to the energy released in order to liberate the electron. Therefore, the energy of the photon can be given by:
E = ΔE = 0 eV = 0 J
Now, we can use the energy-wavelength relationship for photons, which is given by the formula:
E = h·c / λ
Where E is the energy, h is the Planck's constant (6.626 x 10^(-34) J·s), c is the speed of light in a vacuum (3 x 10^8 m/s), and λ is the wavelength of the photon. Rearranging the equation, we can solve for the wavelength of the photon:
λ = h·c / E
Substituting the energy (0 J) into the equation, we get:
λ = (6.626 x 10^(-34) J·s)·(3 x 10^8 m/s) / (0 J)
Since the numerator is not zero and the denominator is zero, we have an indeterminate form: 0/0. This indicates that the energy of the photon is extremely small or negligible compared to the energy levels of the hydrogen atom. Therefore, we can conclude that the wavelength of the photon is undefined or unknown.
In this scenario, it is important to note that the given wavelength (4.02 x 10^(-9) m) corresponds to the particle (electron) emitted after the interaction with the high energy photon, and not the wavelength of the initial photon.