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

Step-by-step explanation:

`E=(hc)/(\lambda)`

E= energy of radiation
`=10.0 kev= 1.6* 10^(-15)J`

`1kev=1.6* 10^(-16)Joules`

h = Planck's constant=
`6.63* 10^(-34)Js`

c = velocity of light
`=3.08* 10^(8)ms^(-1)`

`\lambda` = wavelength of radiation = ?

`\lambda=(hc)/(E)`

`\lambda=(6.63* 10^(-34)Js* 3.08* 10^(8)ms^(-1))/(1.6* 10^(-15)J)`

`\lambda=12.76* 10^(-11)m=0.1270* 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