Electrons in an x-ray tube are accelerated through a potential difference of 3000V. If these electrons were slowed down in a target, what would be the minimum wavelength of the x-ray produce? (Ans: 4.14 x 10⁻¹⁰ m )


Given:


Voltage = V = 3000 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:


Minimum Wavelength = `\λ_{min}` = ?


Solution:

We know that 

`\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 the 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}{3000 V ｘ 1.6 ｘ 10⁻¹⁹ C }`

`\λ_{min}` = `\frac {19.86 ｘ 10⁻²⁶ m³ kg s⁻² }{4800 ｘ 10⁻¹⁹ VC}`

`\λ_{min}` =  0.0041375 ï½˜ 10⁻⁷ V

`\λ_{min}` =  4.1375 ï½˜ 10⁻¹⁰ V  --------------Ans.



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