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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)

http://www.jiskha.com/display.cgi?id=1339073164

Here is the last of the answers by Elena:

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.

• Enough bashing the Tungsten target !! -

it was wrong !

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