A line in the Lyman series of the hydrogen atom emission results from the transition of an
electron from the n=3 level to the ground state level.
a) What n level (#) is the ground state level for the hydrogen atom?
b) What is the energy of this photon in Joules?
c) What is the frequency (in Hz) and wavelength (in nm) of the emitted photon?
*I don't understand this at all help is greatly appreciated*
Bob Pursley answered part of this for you yesterday. It would help if you told us what is confusing you instead of "I don't understand". Basically, the H spectrum consists of lines generated by electrons as they first absorb energy, are promoted to a higher energy level (a higher orbit), then fall back to a lower level. When they fall back to the lower level they give up the energy they absorbed to get to the higher level and this energy falls in the electromagnetic spectrum in the UV, visible, and infrared regions. Historically, these series have been named after people who discovered them; i.e., the Lyman series ENDS with the electron falling to orbit #1 (from orbits 2,3,4,5,etc). That is An electron falling from n = 5 to n = 1 will generate a line at some wavelength and an electron falling form n = 4 to n = 1 will generate another line at some different wavelength. Nothing prevents that electron in n = 5 from falling first to n = 4, then to n = 3, then to n = 2, then to n = 1 and generating four lines in the process. So lines formed when the end orbit is 1 are called the Lyman series (from someone named Lyman). Other series are the Balmer series (where the end level is n = 2, the Paschen series ending in n = 3, Brackett series ending in n = 4 etc).
You now the ground state for the electron is in the n = 1 orbit.
E = 2.180E-18 J x (1/n1^2 - 1/n2^2) is the energy level when an electron falls from n = 2, wherever that may be, to n = 1. Since the Lyman series ends in n = 1, this may be shortened to E = 2.180E-18(1 - 1/n^2). Susbtitute 3 for n. square it and solve for E (in Joules).
E = hc/wavelength will convert from E to wavelength. (or E = h*freq to get freq directly).
Frequency and wavelength are related by
c = freq x wavelength where c is the speed of light (in m/s), freq is in Hz and wavelength is in m.
Sure! I'll break down each part of the question to help you understand it better.
a) The ground state level for the hydrogen atom refers to the lowest energy level that an electron can occupy. In the case of the hydrogen atom, it is represented by the principal quantum number (n) equal to 1. So, the ground state level for the hydrogen atom is n = 1.
b) To find the energy of the photon emitted, we need to use the energy formula given by the equation:
Energy = (hc) / λ
where:
- h is the Planck's constant (which is approximately 6.626 x 10^-34 J·s),
- c is the speed of light in a vacuum (which is approximately 3.00 x 10^8 m/s),
- λ is the wavelength of the emitted photon in meters.
To calculate the energy, we need to know the wavelength of the emitted photon, which we will determine in the next part of the question.
c) To find the frequency (in Hz) and wavelength (in nm) of the emitted photon, we can use the Rydberg formula:
1 / λ = R * (1/n_f^2 - 1/n_i^2)
where:
- λ is the wavelength of the photon in meters,
- R is the Rydberg constant (which is approximately 1.097 x 10^7 m^-1),
- n_f is the final energy level (n=1 for the ground state), and
- n_i is the initial energy level (n=3 for this case).
To find the frequency (in Hz) of the emitted photon, we can use the equation:
c = λ * ν
where:
- c is the speed of light in a vacuum,
- λ is the wavelength of the emitted photon in meters, and
- ν is the frequency of the emitted photon.
To convert the wavelength from meters to nanometers, you can use the conversion factor:
1 nm = 1 x 10^-9 m
I hope this explanation helps. Let me know if you have any further questions!
I'm here to help! Let's break it down step by step.
a) The ground state level for the hydrogen atom is the lowest energy level, where the electron is closest to the nucleus. In this case, the electron is transitioning from the n=3 level to the ground state level. So, the ground state level corresponds to n=1.
b) To find the energy of the photon, you can use the Rydberg formula:
E = ( - 13.6 eV ) * [1/n1^2 - 1/n2^2]
Here, n1 is the initial energy level (n=3 in this case), and n2 is the final energy level, which is the ground state level (n=1).
Plugging in the values:
E = ( - 13.6 eV ) * [1/1^2 - 1/3^2]
E = ( - 13.6 eV ) * [1 - 1/9]
E = ( - 13.6 eV ) * [8/9]
To convert the energy to joules, we know that 1 eV is equal to 1.6 x 10^-19 J. So:
E = ( - 13.6 eV ) * [8/9] * (1.6 x 10^-19 J/eV)
Calculate this value to find the energy of the photon in joules.
c) Now, let's find the frequency and wavelength of the emitted photon. The energy of a photon is related to its frequency (f) and wavelength (λ) through the equation:
E = hf = (hc)/λ
Here, h is the Planck's constant (6.626 x 10^-34 J.s), c is the speed of light (3.0 x 10^8 m/s), and λ is the wavelength.
Rearranging the equation to solve for the frequency:
f = E/h
Find the frequency (in Hz) by dividing the energy E by Planck's constant h.
Then, to find the wavelength, you can rearrange the equation as:
λ = c/f
Find the wavelength (in nm) by dividing the speed of light c by the frequency f.
I hope this helps! Let me know if you have any further questions.