-2
$\begingroup$

Look up the range of wavelengths in nm in the IR region of the electromagnetic spectrum. Convert the lowest IR wavelength to energy in J, frequency in Hz, and wavenumbers in cm-1.

I think I’m going at this the wrong way but if anyone can explain or help with steps that would be really helpful.

$$ E = \frac{(6.676 \times 10^{34})(\pu{3.0 \times 10^8 m/s})}{(\pu{700 \times 10^{-9} nm})} = \pu{0.28 \times 10^{-19} J} $$

$$ V = \frac{\pu{0.28 \times 10^{-19} J}}{6.626 \times 10^{-34}}= 4.28 \times 10^{13} $$

$\endgroup$
0
$\begingroup$

Well, you took $\lambda=700nm$ which is correct.

if anyone can explain or help with step

There isn't really any explanation involved here. The question is very straightforward. Recall these three formulae from your chemistry lectures:

$$\text{Energy}=hv=\frac{hc}{\lambda}$$ and$$\text{wave number}=1/\lambda$$

and $$\text{frequency}=c/\lambda$$

Substitute the necessary values of the constants in them, and now you are good to go!

PS: $1nm=10^{-9}m\neq10^{-9}nm$! Be careful with how you inter convert the units.

Hope it helps!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.