A helium ion He+ emits an ultraviolet photon of wavelength 164 nm. Part A Determ

ID: 1485494 • Letter: A

Question

A helium ion He+ emits an ultraviolet photon of wavelength 164 nm.

Part A

Determine the quantum number of the ion's initial state.

Express your answer as an integer.

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Part B

Determine the quantum number of the ion's final state.

Express your answer as an integer.

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A helium ion He+ emits an ultraviolet photon of wavelength 164 nm.

Part A

Determine the quantum number of the ion's initial state.

Express your answer as an integer.

n1 =

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Part B

Determine the quantum number of the ion's final state.

Express your answer as an integer.

n2 =

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En = -13.6*Z^2/n^2

for He z=2
En = -13.6*2^2/n^2
En = -54.4/n^2

Energy emiited = h*c/lambda
= (6.626*10^-34)*(3*10^8) / (164*10^-9)
= 1.212*10^-18 J
= (1.212*10^-18 ) / (1.6*10^-19 ) eV
= 7.58 eV

En = -54.4/n^2
emergy emitted= 54.4* [1/nf^2 - 1/ni^2]
7.58 = 54.4* [1/nf^2 - 1/ni^2]
[1/nf^2 - 1/ni^2] = 0.139

Now we need to do trial and error:
if ni=2 and nf =1
1/nf^2 - 1/ni^2 = 1/1 - 1/4 = 3/4 = 0.75 eV

if ni=3 and nf =1
1/nf^2 - 1/ni^2 = 1/1 - 1/9 = 0.89 eV

if ni=3 and nf =2
1/nf^2 - 1/ni^2 = 1/4 - 1/9 = 0.139 eV

It matched

A)
Answer: 3

B)
Answer: 2

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