Question

A. An electron emits a photon with an energy of 424 nm as it relaxes to (n=2) what is the initial energy level of this electron?

B. Is energy released or absorbed a result of this transition? Briefly explain.

C. Calculate the wavelength (nm) of the electron having the kinetic energy (K.E. = 1/2 mv^2) of the above transition 434 nm.

D. If an electron has the following three quantum number, n=2, l= 1, m= 0, and m=-1/2. Briefly explain in words what each quantum number means for that electron and sketch the orbital for that electron.

E. Can the location of the electrons in an atom be determined? Discussion must include relevant theroies and princples.

Answer #1

Here we can apply the rydberg equation

lambda is the wavelenght, n is energy level and R is rydberg equation

we have 424 nm are 4.24 x 10^{-7} m

R is 1.097 x10^{7}

^{1 / lambda = 1 /} 4.24 x 10^{-7} =
2358490.5

this value divided by R is 0.21499

0.21499 = 1/n1^{2} - 1/n2^{2}

n2 = 2

1/n2 = 1/2 = 0.5

0.5^{2} = 0.25

0.21499 = 1/n1^{2} - 0.25

1/n1^{2} = 0.21499 + 0.25 = 0.46499

1/ 0.46499 = 2.15 = n^{2}

n = 1.46

n must be equal to 1

b) lets apply the next equation

E = 2.18 x10^{-18} * (1/n2^{2} -
1/n1^{2})

E = 2.18 x10^{-18} * (1 / 1 - 1 / 4)

E = 2.18 x10^{-18} * (1 - 0.25)

E = 1.635 x 10^{-18}

1.635 x 10^{-18} J are absorbed because the electron
goes from n =1 to n =2

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