The energy difference between states A and B is twice the energy difference between states B and C (C > B > A). In a transition (quantum jump) from C to B, an electron emits a photon of wavelength 300 nm.
What is the wavelength emitted when the photon jumps from B to A?
What is the wavelength emitted when it jumps from C to A?
To determine the wavelength emitted when the photon jumps from B to A or C to A, we need to use the energy difference between the states and the relationship between energy and wavelength in quantum mechanics.
Let's denote the energy difference between states A and B as ΔE(AB) and the energy difference between states B and C as ΔE(BC). Given that ΔE(AB) is twice ΔE(BC), we can express it as ΔE(AB) = 2ΔE(BC).
We know that the energy of a photon is given by E = hc/λ, where E is the photon's energy, h is Planck's constant (6.626 x 10^(-34) J·s), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength of the photon.
For a transition from C to B, the energy of the emitted photon is equal to the energy difference between the two states, so we have E(BC_to_B) = ΔE(BC).
Let's now solve for the wavelength emitted when the photon jumps from C to B:
E(BC_to_B) = ΔE(BC)
hc/λ(BC_to_B) = ΔE(BC) [equation 1]
We can rearrange equation 1 to solve for λ(BC_to_B):
λ(BC_to_B) = hc/ΔE(BC)
Given that the wavelength emitted when the photon jumps from C to B is 300 nm, we have:
λ(BC_to_B) = 300 nm = 300 x 10^(-9) m
Now, to find the wavelength emitted when the photon jumps from B to A, we can use the relationship between energy and wavelength once again:
E(AB_to_B) = ΔE(AB)
Using the same equation as before, we solve for λ(AB_to_B):
λ(AB_to_B) = hc/ΔE(AB)
Since the energy difference ΔE(AB) is twice ΔE(BC), we have ΔE(AB) = 2ΔE(BC). Substituting this into the equation above, we get:
λ(AB_to_B) = hc/2ΔE(BC) [equation 2]
Since we know the values of h, c, and λ(BC_to_B) from the previous calculations, we can substitute them into equation 2 to find the wavelength emitted when the photon jumps from B to A.