This problem is killing me [calculation of electron transition]

[eq123]

this problem is killing me! [calculation of electron transition]

i've been trying to solve this problem.. the answer should be 7.. my answer is 0.7 !

What is the value of n_i for an electron that emits a photon of wavelength 93.14 nm when it returns to the ground state in the H atom?

my solution..

n_f=1
λ=93.14 nm×(10^(-9) m)/(1 nm)=93.14×10^(-9) m
∆E=hν=hc/λ=(6.63×10^(-34)×3.00×10^8)/(93.14×10^(-9))≈2.14×10^(-18)
∆E=-2.18×10^(-18) (1/(n_f^2 )-1/(n_i^2 ))
2.14×10^(-18)=-2.18×10^(-18) (1/1^2 -1/(n_i^2 ))
-2.14/2.18=1-1/(n_i^2 )
1.982= 1/(n_i^2 )
n_i^2=0.5045
n_i=√0.5045≈0.7

n_i = n initial
n_f = n final
λ = wavelength ( lambda )
∆E = energy of the transition
h = plank's constant
ν = frequency ( nu )
c = speed of light

if reading the solution is an issue.. just past it in microsoft word and activate the equation mode..

 

[Borek]

if reading the solution is an issue.. just past it in microsoft word and activate the equation mode..

We have something called LaTeX for such situations. And I don't have, use, nor want to use Microsoft Word.

When I solve

$$\frac 1 {93.14nm} = R_{\infty} (\frac 1 {1^2} - \frac 1 {n^2})$$

for n, I get 6.85 - close enough to 7. No idea what is 2.18x10^-18, as far as I can tell Rydberg constant is 1.097x10⁷ m^-1.

--
methods

 

[Cesium]

for n, I get 6.85 - close enough to 7. No idea what is 2.18x10^-18, as far as I can tell Rydberg constant is 1.097x10⁷ m^-1.

2.18x10^-18 is the Rhydberg constant in joules.

 

[Borek]

OK, I know what have happened. Check your math, you made a simple error when solving.

--
methods

 

[eq123]

i already figured it out.. since the electron is returning to its ground state.. it is releasing energy.. which means that ∆E is negative.. after taking that into consideration.. the answer should be 7..

2.18x10-18 is the Rhydberg constant in joules.

wow.. it makes more sense to me now!

thank you.