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 = λmin = ?


Solution:

We know that 

fmax = cλmin


or 

λmin =cfmax

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

λmin` =hcVe


by putting values

λmin =6.62ｘ10⁻³⁴m²kgs⁻¹ｘ3ｘ10⁸ms⁻¹3000Vｘ1.6ｘ10⁻¹⁹C

λmin = 19.86ｘ10⁻²⁶m³kgs⁻²4800ｘ10⁻¹⁹VC

λmin =  0.0041375 ｘ 10⁻⁷ V

λmin =  4.1375 ｘ 10⁻¹⁰ V  --------------Ans.



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