joho Posted November 28, 2006 Report Share Posted November 28, 2006 I need to calculate the wavelength in nm of a photon whose energy is 3.312 keV. Now I know its E= hc/ λ c = 2.997 x 10 8ms -1 (how do I chnge to nm? I know its easy but just can't grasp it!:) )h = 6.626 x 10 -34Js or 4.135 x 10-15 eV.sE= 3312 eV Then how do I rework the equation to get λ?:naughty: Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted November 28, 2006 Report Share Posted November 28, 2006 ::Moved to Science projects and homework:: [math]E=hf[/math] and [math]c=f\lambda[/math] rearranging [math]E=\frac{hc}{\lambda}[/math][math]\lambda=\frac{hc}{E}[/math] Quote Link to comment Share on other sites More sharing options...

ronthepon Posted November 28, 2006 Report Share Posted November 28, 2006 Instead of trying to change the value of [math]c[/math] so as to have it in nanometers(I'll tell you how later), I'd suggest that you convert all quantities into SI units, then convert the final answer to whatever unit you want. This method will help you keep things simple for the tough problems you may encounter. Now how to convert the value of [math]c[/math] to [math]nm/s[/math]? First understand that [math]c[/math] is having the units of velocity, of the form 'the body moves __ distance in __ time'. When the unit is [math]2.997 \times 10^8 m/s[/math] it means that light travels [math]2.997 \times 10^8[/math]meters in one second. You want to see the velocity in [math]nm/s[/math], that means that you want to find out how many nanometers it moves in one second. So simply convert the number of meters into nanometer units. One meter has [math]10^9[/math] nanometers, so it becomes [math]2.997 \times 10^8 \times 10^9 = 2.977 \times 10^{17}[/math] nanometers per second. Jay-qu's demonstrated how to alter equations to have them as per your utility, use his method for the rest of the question. Now, as a thought experiment, or maybe just to test how much you've understood, I ask you to tell me the speed of light in meters per nanosecond. Can you do it? Quote Link to comment Share on other sites More sharing options...

joho Posted November 28, 2006 Author Report Share Posted November 28, 2006 Ronthepon - No can't do the meters per nanosecond thing. Am sure its really obvious though! Just not to me today! Am thinking about it though - might come back with something later!!! This is my working so far: E = hc/λ E = hc/λ E = 3.312 keV or 3.312 x 103 eV or 5.2992 x 10-16 Jh = Planck’s Constant, 6.626 x 10-34 J.s c = velocity of light, 2.997 x 108 m.s λ = wavelength (required) so, λ = hc/ Eλ = 6.626 x 10-34 X 2.997 x 108 / 5.2992 x 10-16 = 3.747 x 10-15 m x 109 = 0.000037473 or 3.7473 x 10-5 nm Having calculated λ above I then need to calculate the angle θ at which you would set a WD spectrometer containing a quartz (SiO2) dispersing crystal with an effective d spacing of 3.343 Ǻ. This is my working for this:nλ = 2 d sin θ n = 1d = 3.343 Ǻ or 3.343 x 10-10 mλ = 3.7473 x 10-5 nm or 3.7473 x 10-14 m so, θ = sin-1 (nλ/2d) = sin-1 (1 x 3.7473 x 10-14 m / 2 x 3.343 x 10-10 m) θ = ??? But now I can't figure out how to get θ!:lol: Quote Link to comment Share on other sites More sharing options...

Jay-qu Posted November 28, 2006 Report Share Posted November 28, 2006 thats simple number crunching now [math]\theta=sin^{-1}(\frac{3.7473*10^{-14}}{2*3.343*10^{-10}})[/math] [math]\theta=sin^{-1}(5.605*10^{-5})[/math] Quote Link to comment Share on other sites More sharing options...

joho Posted November 29, 2006 Author Report Share Posted November 29, 2006 But that gives me an angle of 0.0032 DEG, I thought I must be doing something wrong to get that figure, but I can't see any errors in my calcs! Thank you Jay-qu and Ronthepon for you help! Edit: The moral of this story is to check and check again and check again just to be certain! I had made a mistake in the first calculation of λ which obviously roled through. My angle is 32° :esheriff: Thank you again for your help! Quote Link to comment Share on other sites More sharing options...

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