GR 8677927796770177 | # Login | Register

GR9277 #80
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\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

1. 0.1 angstroms
2. 0.5 angstroms
3. 2 angstroms
4. 25 angstroms
5. 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 $E=hf = h c/\lambda$.

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
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
 omidkhakpour2014-02-03 09:54:59 yo think this wrong but q = e, so we get straight back to electron volts.
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
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...

 itorsics2011-09-11 00:29:31 And $q = e$, so we get straight back to electron volts...
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.
 nyuko2009-10-30 21:59:16 I think memorizing $hc=1240eVnm$ is good enough
 apr20102010-04-08 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.
 gravity2010-11-09 05:12:06 I think memorizing $hc = 5/4 KeV nm$ is good enough
 neon372010-11-12 06:21:21 try 1000 eVnm even easier.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$