A tungsten target is struck by electrons that have been accelerated from rest through a 40KV potential difference. Find the shortest wavelength of the bremsstrahlung radiation emitted? (Ans: 0.31 x 10⁻¹⁰ m)


Given:


Voltage = V = 40 kV = 40 x 10³ V

∴ Plank's Constant = h =  6.62 x 10⁻³⁴ m² kg s⁻¹

∴ Charge on the electron = e = 1.6 x 10⁻¹⁹ C

∴ Velocity of light  = c = 3 x 10 m s⁻¹



To Find:


Shortest Wavelength = λmin= ?


Solution:

We know that (from v=fλ)

fmaxcλmin


or 

λmin =cfmax

where Since Emax = h fmaxfmax = Emaxh  and where Emax = K.E. = Ve (according to law of conservation of energy) So,

λmin =hcVe


by putting values

λmin =`\frac {6.62 x 10⁻³⁴ m² kg s⁻¹ x 3 x 10⁸ m s⁻¹}{40 x 10³ V x 1.6 x 10⁻¹⁹ C }`

λmin = 19.8610²m³kgs²6410¹ VC

λmin =  0.3103125 x 10⁻¹⁰ V

λmin =  0.31 x 10⁻¹⁰ V  --------------Ans.



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