Author Topic: Need some help here... (Read 481 times) Tweet Share

Ahin · « **on:** September 29, 2013, 11:31:14 pm »

Hey guys

I have been staring at this question for hours, and I still have no idea how VCAA end up with this answer...

VCAA physics 2010 exam 2, Area 2, Question 10.

I ended up with 2.55 something... which is close, but not correct...

Can someone please walk me through this question?

Thanks...

SocialRhubarb · « **Reply #1 on:** September 30, 2013, 12:45:22 pm »

Okay, let's start off with some manipulation of formulae.

For an electron:

$mv=\sqrt{2m(\frac{1}{2}mv^2)}$

$p=\sqrt{2mE_k}$

$\lambda=\frac{h}{p}$

For the x-ray:

$E=hf=\frac{hc}{\lambda}=pc\ (\text{if we assume the x-ray and electron have the same wavelength)}$

Now it's only a matter of substituting in all the information which we have.

$E_k=600\ \text{eV}=600\times1.6\times10^{-19}\text{C V}=9.6\times10^{-17}\text{ J}$

$p=\sqrt{2mE_k}=\sqrt{2(9.1\times10^{-31}\text{ kg})(9.6\times10^{-17}\text{ J})}=1.322\times10^{-23}\text{ N s}$

$E=pc=1.322\times10^{-23}\text{ N s}\times3.0\times10^8\text{ ms}^{-1}=3.966\times10^{-15}\text{ J}=\frac{3.966\times10^{-15}}{1.6\times10^{-19}}\text{ eV}\approx 2.48\times10^4\text{ eV}$

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Author Topic: Need some help here... (Read 481 times) Tweet Share

Ahin

Need some help here...

SocialRhubarb

Re: Need some help here...

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