Question

Inorganic chem q1

Answer 1

here is my attempt. I manually calculate all the frequencies first each energy level jump and use the debroglie eqn to find the wavelength, c wasn't given in the question though, i am not sure if it is necessary.

\[\nu _{n1+n2}=Z^2\frac{ R }{ h }\left[ \frac{ 1 }{ n _{1}^2 }-\frac{ 1 }{ n _{2}^2 } \right]\]
\[\nu _{n1+n2}=8^2\frac{ 2.180\times10^{-18} }{ 6.626\times10^{-34} }\left[ \frac{ 1 }{ 1^2 }-\frac{ 1 }{2^2 }\right]= 1.58\times10^{17} \]
\[\nu _{n1+n3}=8^2\frac{ 2.180\times10^{-18} }{ 6.626\times10^{-34} }\left[ \frac{ 1 }{ 1^2 }-\frac{ 1 }{3^2 }\right]= 1.87\times10^{17} \]
\[\nu _{n1+n4}=8^2\frac{ 2.180\times10^{-18} }{ 6.626\times10^{-34} }\left[ \frac{ 1 }{ 1^2 }-\frac{ 1 }{4^2 }\right]= 1.97\times10^{17} \]
\[E=hv=\frac{ hc }{ \lambda } \]
\[\lambda _{n1+n2}=\frac{ 6.626\times10^{-34}\times3\times10^{8} }{ 6.626x10^{-34}\times1.579x10^{17} } = 1.90\times10^{-9}\]
\[\lambda _{n1+n3}=1.60\times10^{-9}\]
\[\lambda _{n1+n4}=1.52x10^{-9}\]

I found that the wavelegth for each jump are all 10^-9 which is within the range between xray and ultraviolet. I have got a feeling that it belongs to ultraviolet, can anyone confirm with me if what i did is correct? any help is appreciated(:

Answer 2

it looks right. \(\checkmark\) assuming all the units are in meters. Also, the equation you wrote is not de Broglie's, it's Planck's. hmm also since you're dealing with photons you could've converted \(\nu\) to \(\lambda\) using "\(c=\nu*\lambda\)", but it's ultimately the same thing.

Answer 3

@aaronq lookin' good :P

Answer 4

thank you! :D ill note of that equation!

Answer 5

lmao it's some girl from a band @abb0t i know she's eye candy