What is the energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2?
E = R(1/n^2 - 1/n^2)
R = 1.097E7*(1/4 - 1/25)
To determine the energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2, we can use the formula:
ΔE = E_final - E_initial
where ΔE is the change in energy, E_final is the energy of the final state, and E_initial is the energy of the initial state.
In hydrogen, the energy levels are given by the formula:
En = -13.6 eV / n^2
where n is the principal quantum number.
First, let's calculate the energy of the electron in the final state, E_final, using n = 2:
E_final = -13.6 eV / (2^2)
= -13.6 eV / 4
= -3.4 eV
Next, let's calculate the energy of the electron in the initial state, E_initial, using n = 5:
E_initial = -13.6 eV / (5^2)
= -13.6 eV / 25
= -0.544 eV
Now, we can find the change in energy, ΔE:
ΔE = E_final - E_initial
= -3.4 eV - (-0.544 eV)
= -2.856 eV
Finally, to convert the energy to joules, we can use the conversion factor:
1 eV = 1.602 x 10^-19 Joules
Thus, to convert the energy from eV to Joules:
Energy = ΔE x (1.602 x 10^-19)
= -2.856 eV x (1.602 x 10^-19)
= -4.58 x 10^-19 Joules
Therefore, the energy of the photon emitted when the electron in a hydrogen atom drops from energy level E5 to energy level E2 is approximately -4.58 x 10^-19 Joules. Note that the negative sign indicates that energy is being emitted.