The electron from a hydrogen atom drops from an excited state into the ground state. When an electron drops into a lower-energy orbital, energy is released in the form of electromagnetic radiation.
Part A: How much energy does the electron have initially in the n=4 excited state?
answer I got = -1.36*10^-19 J
Part B: If the electron from Part A now drops to the ground state, how much energy is released?
the answer i got was -2.04*10^-18 J
but i don't understand part :
What is the wavelength lambda of the photon that has been released in Part B?
can someone please explain this one to me, thank you
I didn't check your answers to A and B.
For the part you don't understand, subtract the energy of the electron in the two levels; that is the energy released. Then substitute into E = hc/wavelength and solve for wavelength.The unit for wavelength in this equation is meter.
When an electron moves from the n=2 to n=4 state, Does it gain or lose energy? In a more to n=4, is a photon emitted or absorbed?
To determine the wavelength (λ) of the photon released when an electron drops to the ground state, you can use the relationship between energy (E) and wavelength (λ) given by the equation:
E = hc/λ
where h is the Planck's constant (6.626 x 10^-34 J·s) and c is the speed of light (3.0 x 10^8 m/s).
We know the energy released in Part B is -2.04 x 10^-18 J. Plugging this value into the equation, we can solve for the wavelength:
-2.04 x 10^-18 J = (6.626 x 10^-34 J·s)(3.0 x 10^8 m/s) / λ
Rearranging the equation to solve for λ:
λ = (6.626 x 10^-34 J·s)(3.0 x 10^8 m/s) / (-2.04 x 10^-18 J)
Now, let's calculate the value using these numbers:
λ = (6.626 x 10^-34 J·s)(3.0 x 10^8 m/s) / (-2.04 x 10^-18 J)
λ ≈ 9.79 x 10^-7 m
So, the wavelength (λ) of the photon released when the electron drops to the ground state is approximately 9.79 x 10^-7 meters (or 979 nanometers).