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Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated by such beam directed head-on into a lead wall? express your answer numerically in nanometers.

Respuesta :

shinmin
To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength 
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV 

= 1.240x10^-10 m 

= 1.240x10^-1 nm

Answer: 0.1276 nm

Explanation:

[tex]E=\frac{hc}{\lambda}[/tex]

E= energy of radiation [tex]=10.0 kev= 1.6\times 10^{-15}J[/tex]          

[tex]1kev=1.6\times 10^{-16}Joules[/tex]

h = Planck's constant= [tex]6.63\times 10^{-34}Js[/tex]

c = velocity of light [tex]=3.08\times 10^{8}ms^{-1}[/tex]

[tex]\lambda[/tex] = wavelength of radiation = ?

[tex]\lambda=\frac{hc}{E}[/tex]

[tex]\lambda=\frac{6.63\times 10^{-34}Js\times 3.08\times 10^{8}ms^{-1}}{1.6\times 10^{-15}J}[/tex]

[tex]\lambda=12.76\times 10^{-11}m=0.1270\times 10^{-9}m[/tex]

[tex]\lambda=0.1276nm[/tex]

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