please write out all of the steps you took and the final answer so i can study f

Question

please write out all of the steps you took and the final answer so i can study from it!! thank you for your help!! i really appreciate it!!

Consider a Bohr model of doubly ionized lithium.

(a) Write an expression similar to En = -13.6 eV/n2 E2 = -13.6 eV/22 = -3.40 eV for the energy levels of the sole remaining electron. (Use the following as necessary: n.)

En = ? Ev

(b) Find the energy corresponding to n = 6.
2 = ? eV

(c) Find the energy corresponding to n = 2.
3= ? eV

(d) Calculate the energy of the photon emitted when the electron transits from the sixth energy level to the second energy level. Express the answer both in electron volts and in joules.
4= ? eV
5= ? J

(e) Find the frequency and wavelength of the emitted photon.
6= ? Hz
7= ? nm

(f) In what part of the spectrum is the emitted light?

choose one: x-ray region... gamma region... ultraviolet region... infrared region... visible light region

thank you!!!

Explanation / Answer

1) For the bohr model, we use the following assumptions. 1) The electrons move in circular orbits 2) Angular momentum is quantized as some multiple of h-bar (h/2pi) If you draw out the electron orbiting the nucleus, you get that the force is given by coulombs law, but that it can also be written in terms of a centripetal force as follows F = kQq/r^2 = m v^2 / r Where Q is the charge of the nucleus, q is that of an electron, m is the mass of an electron, r is the orbit radius, and v is the orbit velocity, and k is 8.99*10^9 multiply both sides by r^2 to get the following kQq = m v^2 r = (mvr)v Remember that our second assumption assumes that angular momentum (L) is some integer quantity of h-bar, so mvr = n*h-bar where n is some integer, so lets replace mvr with n*h-bar to get the following. kQq = n*h-bar*v Now solve for v to get the following expression v = kQq/n*h-bar Now we can solve for the kinetic energy of the particle using the classical formula K=1/2 mv^2 , plug in our formula for v to get the following K = (1/2)*m*(kQq/n*h-bar)^2 This is the Kinetic energy, now we need to solve for the potential energy, because total energy equals kinetic plus potential. The potential energy between two charges is given by -kQq/r , if we return to the equation we had earlier (kQq = mv^2 *r) then we can solve for r as follows by dividing mv^2 from both sides r= kQq / mv^2 Remember however, that 1/2 mv^2 equals the kinetic energy, so we can replace the denominator with 2K so that r=kQq/2K So now if we plug this value for r into our equation for potential energy we get: U = -kQq / (kQq/2K) The kQq cancels and we get U = -2K Now we are ready to combine our kinetic and potential energy, E = K + U E = K -2K E = -K So it is clear that the total energy is just negative the kinetic energy we solved earlier, so the total energy is given by the following: E = - (1/2)*m*(kQq/n*h-bar)^2 Notice that if you plug in the charge of 1 proton for the charge of the nucleus and you get the formula for hydrogen energy levels, E = -13.6/n^2, however, for lithium you simply change Q to be the charge of 3 protons, plugging this in along with the other constants k,q, and h-bar you get the following formula: E = -122.49 / n^2 where this energy is in electron volts *** Note that a negative energy is what we would expect for a bound sytem such as this because energy would be required to "break" the electron nucleus bond 2/3 ) To find the energy for the n=6 and n=2 levels we plug in 6 and 2 for n to our energy equation E6 = -122.49 / 6^2 = -122.49/36 = -3.4 eV E2 = -122.49 / 2^2 = -122.49/4 = -30.6 eV 4) If an electron transitions from the 6th to the 2nd energy level, we use energy conservation to find the energy of the emitted photon E0 = Efinal The initial energy is just the energy of the sixth energy level, the final energy is the energy of the second level plus the energy of the emitted photon E6 = E2 + Ephoton We solved for E6 and E2 in parts 2 and 3, so lets plug that in now -3.4 = -30.6 + Ephoton Solve for Ephoton to get Ephoton = 27.2 eV Converting this to joules we get Ephoton = 4.36E-18 J 5) Use the formula for the energy and wavelength of a photon E = hc/lambda = hf, rearrange these to solve for lambda and frequency in terms of energy, h, anc c to get lambda = hc/E f = E/h plug in the values of h, c, and E to get the wavelength and frequency, the values you should get are: Wavelength (lambda) = 45.59 nm Frequency = 6.58E15 Hz = 6.58 PHz 6) This wavelength and frequency correspond to the Ultraviolet portion of the electromagnetic spectrum

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