GREPhysics.NET: GR9277 #80

GR9277 #80

Problem

GREPhysics.NET Official Solution

\prob{80}
A beam of electrons is accelerated through a potential difference of 25kV
in an x-ray tube. The continuous x-ray spectrum emitted by the target of the tube will have a short wavelength limit of most nearly

0.1 angstroms
0.5 angstroms
2 angstroms
25 angstroms
50 angstroms

Quantum Mechanics $\Rightarrow$ }Planck Energy

The key equation is $E=hc/\lambda$ . Since $hc=1.24E-6eVm$ and $E=25E3eV=2.5E4eV$ , one can immediately plug the quantities in to solve for $\lambda=hc/E = 1.24E-6/2.5E4=0.5E-10$ , which is just choice (B).

No knowledge of X-rays required, other than the elementary knowledge that it's a electromagnetic wave and allows one to write the Planck energy as $E=hf = h c/\lambda$ .

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Comments

omidkhakpour
2014-02-03 08:59:51

this quastion is wrong becaus the 25KV is "25K electron volt". The question said it's V = volt, not eV=electron volt

omidkhakpour
2014-02-03 09:54:59

yo think this wrong but q = e, so we get straight back to electron volts.

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omidkhakpour
2014-02-03 08:59:34

this quastion is wrong becaus the 25KV is "25K electron volt". The question said it's V = volt, not eV=electron volt

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ali8
2011-07-13 08:28:32

Sorry but, how did all of you assumed that the 25KV is "25K electron volt". The question said it's V = volt, not eV=electron volt.

So we should use E=qV...

itorsics
2011-09-11 00:29:31

And $q = e$ , so we get straight back to electron volts...

Reply to this comment

gumby
2009-10-29 21:43:39

Memorize this useful quantity! It's faster than trying to multiply h*c all the time.
$h c = 1239.87 eV nm$
Mnemonic: first 3 digits, last 3 digits.

nyuko
2009-10-30 21:59:16

I think memorizing $hc=1240eVnm$ is good enough

apr2010
2010-04-08 09:15:47

On the order of atom lattice constants / near neighbor distances ( $\approx 1 \AA$ ), so it should be 0.5 or smaller. The resolution is in general not as good with x-ray tubes, that's why I have chosen B.

gravity
2010-11-09 05:12:06

I think memorizing $hc = 5/4 KeV nm$ is good enough

neon37
2010-11-12 06:21:21

try 1000 eVnm even easier.

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Sorry but, how did all of you assumed that the 25KV is "25K electron volt". The question said it's V = volt, not eV=electron volt. So we should use E=qV...

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