What is the wavelength of the light?

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SUMMARY

The discussion focuses on calculating the wavelength of light emitted during an electron transition in a one-dimensional box of length 5.2 nm, specifically from the n = 7 state to the n = 6 state. The initial calculation yielded an energy of 2.9 x 10^-20 J and a wavelength of 6.85 x 10^-6 m, which was later identified as incorrect. The correct approach involves using the formula λ = (8(m_e c^2) L^2)/(hc)(1/(n_f^2 - n_i^2)), with constants such as hc = 1.240 eV·μm and m_ec^2 = 511 eV, to ensure accurate results.

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Calculate the wavelength of the light emitted when an electron in a one-dimensional box of length 5.2nm makes a transition from the n = 7 state to the n = 6 state.
Express your answer with the appropriate units.

I calculated in this way:
E = n^2*h^2/8*m*a^2
E= (7^2-6^2)*(6.626*10^-34)^2/8*(9.1*10^-31)*(5.2*10^-9)^2
E= 2.9*10^-20 J

E = hc/λ
λ = 6.85*10^-6 m

But the answer is wrong.

What is wrong with my calculation?
 
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How do you know the answer is wrong?

It can help if you do the algebra first, then use more convenient units:$$\lambda=\frac{8(m_e c^2) L^2}{hc}\frac{1}{n_f^2-n_i^2}$$

##hc=1.240\text{eV$\mu$m}##
##m_ec^2=511\text{eV}##
##L=0.0052\text{$\mu$m}##
... give it a go.
(Don't forget to check my algebra to get that equation.)
 
Simon Bridge said:
How do you know the answer is wrong?

It can help if you do the algebra first, then use more convenient units:$$\lambda=\frac{8(m_e c^2) L^2}{hc}\frac{1}{n_f^2-n_i^2}$$

##hc=1.240\text{eV$\mu$m}##
##m_ec^2=511\text{eV}##
##L=0.0052\text{$\mu$m}##
... give it a go.
(Don't forget to check my algebra to get that equation.)

It is the same calculation as mine.
I did the exercise on MasteringChemistry. It only said my answer is wrong.
 
But I got a different order of magnitude from you.
Repeat the calculation ... you have misplaced a decimal point someplace.
 

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