Electron exhibiting a particle wavelength of 5.2 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
E = hc/wavelength.You don't have units on that electron.
But we supposed to determinate wavelength
Are you saying that electron wavelength is invalid here?
No. The electron emitting the 5.2E-9 (meters?) must have been moved from the ground state to an upper energy level. So the proton must have had that minimum energy. However, I notice on re-reading the question that it speaks of FREE electrons. Perhaps I'm not on the same "wavelenth" (no pun intended) as the problem since I'm not think FREE electrons.
what's result?
sigma-2p
sigma-2p
CN-
right answers
To determine the wavelength of the photon that generated the free electrons, we can utilize the principle of conservation of energy. The energy of the photon is equal to the energy difference between the initial and final states of the hydrogen atom, which can be calculated using the Rydberg formula:
1 / λ = R * (1 / n1^2 - 1 / n2^2)
where:
- λ is the wavelength of the photon
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1)
- n1 is the initial principal quantum number (in this case, since the hydrogen atoms were in the ground state, n1 = 1)
- n2 is the final principal quantum number (in this case, since free electrons are generated, n2 = ∞)
Substituting the values into the formula:
1 / λ = 1.097 × 10^7 m^-1 * (1 / 1^2 - 1 / ∞^2)
To simplify further, we can ignore the 1 / ∞^2 term, as it approaches zero. Thus:
1 / λ ≈ 1.097 × 10^7 m^-1 * (1 / 1^2)
1 / λ ≈ 1.097 × 10^7 m^-1
Taking the reciprocal of both sides:
λ ≈ 1 / (1.097 × 10^7 m^-1)
λ ≈ 9.110 x 10^-8 meters
Therefore, the wavelength of the photon that generated the free electrons is approximately 9.110 x 10^-8 meters or 91.10 nm.