Question

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.

Answer 1

Answer: 0.1276 nm

Explanation:

$E=\frac{hc}{\lambda}$

E= energy of radiation $=10.0 kev= 1.6\times 10^{-15}J$          

$1kev=1.6\times 10^{-16}Joules$

h = Planck's constant= $6.63\times 10^{-34}Js$

c = velocity of light $=3.08\times 10^{8}ms^{-1}$

$\lambda$ = wavelength of radiation = ?

$\lambda=\frac{hc}{E}$

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

$\lambda=12.76\times 10^{-11}m=0.1270\times 10^{-9}m$

$\lambda=0.1276nm$

Answer 2

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