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 ï½˜ 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 10⁸ m s⁻¹



To Find:


Shortest Wavelength = `\λ_{min}`= ?


Solution:

We know that (from v=fλ)

`\f_{max}` = `\frac {c}{λ_{min}}`


or 

`\λ_{min}` =`\frac {c}{f_{max}}`

where Since `\E_{max}` = h `\f_{max}` ⇒ `\f_{max}` = `\frac {E_{max}}{h}`  and where `\E_{max}` = K.E. = Ve (according to law of conservation of energy) So,

`\λ_{min}` =`\frac {hc}{Ve}`


by putting values

`\λ_{min}` =`\frac {6.62 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ 3 ｘ 10⁸ m s⁻¹}{40 ï½˜ 10³ V ï½˜ 1.6 ｘ 10⁻¹⁹ C }`

`\λ_{min}` = `\frac {19.86 ｘ 10⁻²⁶ m³ kg s⁻² }{64 ｘ10⁻¹⁶  V C}`

`\λ_{min}` =  0.3103125 ï½˜ 10⁻¹⁰ V

`\λ_{min}` =  0.31 ｘ 10⁻¹⁰ V  --------------Ans.



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