Electron and wavelength problem

[David112234]

Homework Statement

Electron on n=6 level emits wavelength of 410.2nm.
What energy level does it move too?

Homework Equations

1/L = R ( 1/N^2 - 1/n^2)
where N is initial and n final
R=1.096779*10^7 m

The Attempt at a Solution

1/410.2 = 10967790 (1/36 - 1/n^2)
1/4498975152 = (1/36 - 1/n^2)
-1/36 -1/36
4498975152 =n^2
sqrt
n=67074...
this number is wrong, it is not an energy level, and not one of the answers, what did I do wrong?

 

[DrClaude]

1/L = R ( 1/N^2 - 1/n^2)
where N is initial and n final

Careful here. N is the lower level and n is the upper level, not necessarily initial and final.

Put the units in your equation, you will see where it went wrong.

 

[David112234]

Careful here. N is the lower level and n is the upper level, not necessarily initial and final.

Put the units in your equation, you will see where it went wrong.

alright,
R is in m^-1
and L is in nm,

so it will be .0000004102m*10967760m^-1 = .449897515 m/m
Now I add 36 and take the sqrt, which gives me a number slightly larger than 6, which still is not correct. Did I miss other units?

 

[DrClaude]

so it will be .0000004102m*10967760m^-1 = .449897515 m/m

Almost. I think it is simpler to make calculation using powers of 10 (less chance of an error):
410.2×10^-9 m × 1.096779×10⁷ m^-1 = ?

Now I add 36 and take the sqrt

You can't do that,
$$
\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}
$$
You need to add 1/36 to 1 over the number you will get for λ×R.

 

[David112234]

Almost. I think it is simpler to make calculation using powers of 10 (less chance of an error):
410.2×10^-9 m × 1.096779×10⁷ m^-1 = ?You can't do that,
$$
\frac{1}{a} + \frac{1}{b} \neq \frac{1}{a+b}
$$
You need to add 1/36 to 1 over the number you will get for λ×R.

I thought since all the terms were under 1 that I could just solve the denominators, like this property

X^(2x+5) = x¹⁴
2x+5= 14

also, how do you display fractions on this forum like you did?

 

[DrClaude]

I thought since all the terms were under 1 that I could just solve the denominators, like this property

X^(2x+5) = x¹⁴
2x+5= 14

That makes use of the properties of the logarithm
$$
\begin{align*}
x^a &= x^b \\
\log(x^a) &= \log(x^b) \\
a \log(x) &= b \log(x) \\
a &= b
\end{align*}
$$
It doesn't work for $a^{-1} + b^{-1}$.

also, how do you display fractions on this forum like you did?

I used the Latex capabilities of PF. See https://www.physicsforums.com/help/latexhelp/ for more info.

 

[David112234]

That makes use of the properties of the logarithm
$$
\begin{align*}
x^a &= x^b \\
\log(x^a) &= \log(x^b) \\
a \log(x) &= b \log(x) \\
a &= b
\end{align*}
$$
It doesn't work for $a^{-1} + b^{-1}$.I used the Latex capabilities of PF. See https://www.physicsforums.com/help/latexhelp/ for more info.

Thank you. I did it all out and the value I got it .66666, which still does not seem right

 

[DrClaude]

I did it all out and the value I got it .66666, which still does not seem right

Indeed. I get a nice value. Could you show your calculation?