Asked by Rebecca
A tungsten target is struck by electrons that have been accelerated from rest through a 38.9-kV potential difference. Find the shortest wavelength of the radiation emitted. (answer has to be in nm)
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Answered by
Damon
This question has been answered and answered and answered:
http://www.jiskha.com/display.cgi?id=1339073164
http://www.jiskha.com/display.cgi?id=1339073164
Answered by
Damon
Here is the last of the answers by Elena:
Physics-PLEASE HELP ASAP - Elena, Thursday, June 7, 2012 at 10:41am
For Bremsstrahlung radiation (or "braking X- radiation" )
the low-wavelength cutoff may be determined from Duane–Hunt law :
λ =h•c/e•U,
where h is Planck's constant, c is the speed of light, V is the voltage that the electrons are accelerated through, e is the elementary charge
λ = 6.63•10^-34•3•10^8/1.6•10^-19•3.89•10^4 = 3.19•10^-11 m.
Physics-PLEASE HELP ASAP - Elena, Thursday, June 7, 2012 at 10:41am
For Bremsstrahlung radiation (or "braking X- radiation" )
the low-wavelength cutoff may be determined from Duane–Hunt law :
λ =h•c/e•U,
where h is Planck's constant, c is the speed of light, V is the voltage that the electrons are accelerated through, e is the elementary charge
λ = 6.63•10^-34•3•10^8/1.6•10^-19•3.89•10^4 = 3.19•10^-11 m.
Answered by
Damon
Elena already answered it !
Answered by
Rebecca
it was wrong !
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