Posted on April 12.
From the information below, identify element X.
a. The wavelength of the radio waves sent by an FM station
broadcasting at 97.1 MHz is 30.0 million (3.00 e7) times
greater than the wavelength corresponding to the energy difference
between a particular excited state of the hydrogen
atom and the ground state.
b. Let V represent the principal quantum number for the valence
shell of element X. If an electron in the hydrogen atom
falls from shell V to the inner shell corresponding to the excited
state mentioned above in part a, the wavelength of light
emitted is the same as the wavelength of an electron moving
at a speed of 570. m/s.
c. The number of unpaired electrons for element X in the ground
state is the same as the maximum number of electrons in an
atom that can have the quantum number designations n = 2,
mL = -1, and mS = -1/ 2.
d. Let A equal the charge of the stable ion that would form when
the undiscovered element 120 forms ionic compounds. This
value of A also represents the angular momentum quantum
number for the subshell containing the unpaired electron(s)
for element X.
a.
i used lambda= nu/c
97.1Mhz*10^6s^-1=97,100,000s^-1
lambda= 97,100,000/2.998e8= 0.324m
0.324m/30,000,000 = 1.08e-8m
I then found the energy for n=1 through n=5
n=1 E= -2.18e-18J/atom
n=2 -5.45e-19
n=3 -2.42e-19
n=4 -1.36e-19
n=5 -8.72e-20
DrBob222 said:
The difference between the energy of the orbits (the N values) will be the energy of the electron when it makes the transition between any one of them to any other below it. Convert energy difference to wavelength, delta E = hc/lambda and compare with the wavelength of the radio station divided by 30 million. That will tell you N for the excited state.
How do I do that. Do I do each difference. I really don't understand any of this please help! I understand about getting the difference and the converting to wavelength but the difference of which n's and why.
Thanks for posting again on page 1. I hardly ever go back to page 8 or 9 where your original post is by now.
First, you need to correct the first piece of data you have.
c = nu*lambda, not what you have.
3E-8 meters = 97.1E6Hz*lambda
lambda = 3E-8/97.1E6 = ?? lambda of the radio wave;
and ??/30,000,000 = xx = lambda of the wavelength when the electron makes the transition in the hydrogen atom discussed in part a of the problem.
Then Energy of the electron = E = hc/lambda = hc/xx.
All of your E values for N=1 through N=5 are correct. I would add another one for N=6 just for good measure.
Here is the scoop. An electron falls from one of those upper levels to the ground state (look at the problem--it says it falls to th ground state and the ground state is N=1). The energy it gives off is the difference between the upper state energy and the lower state energy; so, just subtract E2-E1 = energy when electron falls from N=2 to N=1. Then E3-E1 = energy when the electron falls from N=3 to N=1 etc. Convert each of these energies to lambda and compare with lambda from part a of the problem OR compare the energy difference to the energy in part a of the problem. E2-E1 or E3-E1 or E4-E1 or E5-E1 or E6-E1 will match. All of this is for the purpose of identifying the excited state (the upper level or N = ??) because the next part of the problem has the electron falling from some higher level to the excited state so you much know what the excited state is.
Post your work if you get stuck. That helps, you see, for I could tell you that your initial calculation was not correct. Good luck.
now I have:
38e8/97.1e6 = 3.089m
3.09m/30,000,000 = 1.029e-7
E= 6.626e-34*3e8/1.029e-7
= 1.932e-18J
E2-E1=1.635e-18
E3-E1=1.938e-18
E4-E1= 2.044e-18
E5-E1= 2.093e-18
E6-E1= 2.119e-18
so it would be N=3
Absolutely right. The electron is in the N=3 level, the excited state, and falls to the ground state where N=1. Now you are ready to determine V, the principal quantum number in part b.
Remember the De Broglie wavelength, lambda = h/mv.
don't you need a mass to used the de broglie equation or is that what we are solving for?
You have a mass. It's the mass of the electron that is traveling at whatever? its velocity. So you have h, m, and v and you can calculate lambda. And from that you can calculate E = hc/lambda if you wish.
sorry.. i still don't understand the mass... do you use the mass of hydrogen 1.008? In class everything is more clear, but I just don't understand these problems.
b. Let V represent the principal quantum number for the valence
shell of element X. If an electron in the hydrogen atom
falls from shell V to the inner shell corresponding to the excited
state mentioned above in part a, the wavelength of light
emitted is the same as the wavelength of an ELECTRON moving
at a speed of 570. m/s.
The secret is there. An electron moving at a speed of 570 m/s has what wavelength? The De Broglie equation is
lambda = h/mv.
You know h, you know mass of the ELECTRON (I made it all caps from your original question), and you know v. Calculate lambda. And from that, if you wish, you may calculate Energy = E = hc/lambda.
These railroad problem CAN be a little confusing because we are overwhelmed (innundated) with information and it is coming at us from all directions. This particularly problem is more confusing than most because it goes from talking about one thing to another; that is, from 97.1 MHz on the radio to a electron in a hydrogen atom and from a principal quantum number to the same "system" in a hydrogen atom. It gets even worse for part c. Part d is a little better.
well the mass of an electron is 9.109e-31kg, right?
yes
lambda= 6.626e-3/(9.109e-31)(570)= 1.276e25m
E= (6.626e-34)(3e8)/1.276e25
=1.558e-50J
I think h = 6.626E-34 J*s and not E-3. That will make a difference in lambda.
E= ?? also is wrong. I have lambda =- 1.276E-6 m for lambda and 1.558E-19 J for energy. Your number is ok but the exponent is wrong. After you have done that, then start subracting the energy of one level from Energy of level 3 (that's the excited state in part a) of the H atom. Just like the process you did in part a. You will find one of those (E6-E3 or E5-E3 or E4-E3) match the speed of the electron's energy when it is moving 570 m/s. If that doesn't do it, you will need to go back and calculate energy for N=7, N=8 etc but I think one of these will do it.
Check my last post for the A,B, and C problem. I don't buy the final pressure being 1 atm.
yes you're right it wrote it wrong.. now i get what you have
for E5-E3 i get 1.548e-19
for the A,B,C problem I used V=18.60L, maybe I gave the wrong moles
OK. So now you know the principal quantum number of element X is N=5 (or in their terms, V=5). Reread part b of the question so you will understand what we have just done.
i used the 1.4L of A to calc new moles? should i have done that?
I understand the v=5 and part b.
I have class now so I will be gone for a while. Thank you for all of your help. hopefully you will still be available when I come back. Thank you!
p = nRT/V= 0.446*0.08205*273/18.60 = ??. Check my thinking. I don't get 1 atm using either mol A remaining or mols dichloroethane (mols B)formed. The only way I can obtain 1 atm is by using 10.0 L for volume. p = 0.446*0.08205*273/10 = 0.999 which we would round to 1. Also, I don't know how you got p1v1 = p2v2 to come out, either.
ok. Glad you understand part b.
Yes, you should have converted 1.4 L to mols. Also convert 8.6 L of compound B to mols. Add them together for total n. Use 18.60 for volume.
now i get P=0.537 atm
What's the answer to this problem?
I got molybdenum...
To summarize:
a. The wavelength of the radio waves sent by an FM station broadcasting at 97.1 MHz is 1.029e-7 meters.
b. The principal quantum number for the valence shell of element X is V=5.
c. The number of unpaired electrons for element X in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations n=2, mL=-1, and mS=-1/2.
d. The charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds is A.
Please note that part d is still unknown and requires more information.
To identify element X, we need to analyze the given information and solve the equations step-by-step. Let's break it down:
a) The wavelength of the radio waves sent by an FM station broadcasting at 97.1 MHz is 30.0 million times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state.
To calculate the wavelength, use the formula λ = c/ν, where λ represents wavelength, c is the speed of light (2.998 x 10^8 m/s), and ν is the frequency in Hz.
Given that the frequency is 97.1 MHz (97.1 x 10^6 Hz), we can calculate λ as follows:
λ = (2.998 x 10^8 m/s) / (97.1 x 10^6 Hz)
λ ≈ 3.089 m
Now, divide this wavelength by 30 million to find the difference in wavelengths between the radio waves and the energy difference in the hydrogen atom:
(3.089 m) / (30,000,000) ≈ 1.029 x 10^-7 m
Next, convert this energy difference to wavelength. The formula is ΔE = hc/λ, where ΔE is the energy difference, h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light, and λ is the wavelength.
You already calculated the energy difference for different excited states (E2-E1, E3-E1, etc.). Choose the energy difference that matches the value calculated in part a. In this case, E3-E1 is approximately equal to 1.935 x 10^-18 J.
Using the equation ΔE = hc/λ, rearrange it to solve for λ:
λ = hc/ΔE = (6.626 x 10^-34 J·s) × (3 x 10^8 m/s) / (1.935 x 10^-18 J) ≈ 1.276 x 10^-6 m
So, the wavelength of light emitted when the electron falls from shell V to the excited state (N=3) is approximately 1.276 x 10^-6 m.
b) The number of unpaired electrons for element X in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations n = 2, mL = -1, and mS = -1/2.
To find the number of unpaired electrons in the ground state, we need to consider the quantum number designations. According to the maximum number of electrons formula, it is given by 2n^2, where n is the principal quantum number.
By substituting n = 2 into the formula, we find:
2(2^2) = 2(4) = 8
So, element X in the ground state has 8 unpaired electrons.
c) Let A equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of A also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element X.
To find A, refer back to the number of unpaired electrons in the ground state (from part b). Since element X has 8 unpaired electrons, the corresponding angular momentum quantum number (A) will be 8.
Therefore, A represents the angular momentum quantum number for the subshell containing the unpaired electrons in element X.
By analyzing the information and solving step-by-step, we have identified element X with relevant quantum numbers and calculated various values.