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 10⁸ m s⁻¹



To Find:


Shortest Wavelength = λmin= ?


Solution:

We know that (from v=fλ)

fmax = cλmin


or 

λmin =cfmax

where Since Emax = h fmax ⇒ fmax = 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 = 19.86ｘ10⁻²

\λ_{min} =  0.3103125 ｘ 10⁻¹⁰ V

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



Similar Questions:



************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.