Solve Inductance for 550nm EM Waves w/17pF Capacitor

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To determine the required inductance for a 17 pF capacitor in an oscillator generating 550 nm electromagnetic waves, the formula f = 1/(2π√(LC)) was rearranged to isolate L. The calculations show that L equals approximately 5.0 x 10^-21 H, which is an extremely small value. The prefix 'p' in 17 pF stands for pico, equivalent to 10^(-12). While the algebra appears correct, the physical implications of such a small inductance may warrant further discussion. Overall, the calculations and understanding of the components involved seem accurate.
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Hi all, I've managed to solve the following question but the answer I'm getting doesn't seem right...

What inductance must be connected to a 17 pF capacitor in an oscillator capable of generating 550 nm (i.e. visible) electromagnetic waves? Comment on your answer.

To solve it I just rearranged the following equation and isolated L

f=\frac{1}{2\pi\sqrt{LC}}

I'm also not sure what the prefix 'p' in the 17pF stands for...

Thanks
 
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p stands for pico, which is equivalent to 10^(-12).
 
So this is what I have now, the answer I'm getting is extremely small, could someone please just check over my work to see if I have made any errors... Thanks

<br /> \begin{gather*}<br /> C=17pF=1.7\times10^{-11}F\\<br /> \lambda=550nm=5.5\times10^{-7}m\\<br /> c=3.0\times10^8m/s<br /> \end{gather8}<br />

<br /> \begin{gather*}<br /> f=\frac{1}{2\pi\sqrt{LC}\\<br /> \frac{c}{\lambda}\sqrt{LC}=\frac{1}{2\pi}}\\<br /> LC=(\frac{\lambda}{2c\pi})^2\\<br /> L=(\frac{\lambda}{2c\pi})^{2}(\frac{1}{C})\\<br /> L=5.0\times10^{-21}\\<br /> \end{gather*}<br />
 
Your algebra looks right to me. I don't remember much about inductance to comment on the physics of it.
 
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