Question

An Electron is incident on a nitrogen atom (mass=7.37x10^-25kg). The electrons initial velocity is 4x10⁴ m/s. The nitrogen atom is at rest. If The electron deflects backwards at an angle of 35 degrees below the horizontal, and at 3.7x10⁴m/s what is the velocity and direction of the nitrogen atom? (mass of an electron is 9.11x10^-31kg)

Answer 1

Using conservation of momentum, before collision = after collision

   : let the mass of electron be Me and mass of nitrogen atom be Mn

  : let velocity of the electron before collision = Ue (i)    { (i) denotes in positive x-direction }

: let velocity of nitrogen after collision be= Vn (i) + Ln (j) {(i) and (j) denotes diredction in positive x and y direction}

Me*Ue(i) = Me*(3.7*104)*cos35 (-i) +  Me*(3.7*104)*sin35 (-j) + Mn*Vn (i) + Mn*Ln (j)

comparing i direction with i on both sides and j direction with j

Me*Ue = -Me*(3.7*104)*cos35 + Mn*Vn and 0= -Me*(3.7*104)*sin35 + Mn*Ln

9.11*10-31*4*104 + 2.761*10-26 =   Mn*Vn and 1.933*10-26 =  Mn*Ln

3.644*10-26  +  2.761*10-26 =   Mn*Vn and Ln = 0.026 m/s

Vn = 0.0869m/s and Ln = 0.026 m/s

hence velocity of nitrogen atom is 0.0869 (i) + 0.026 (j)

direction = tan-1 =0.026/0.0869 =>  = 16.646 degree from x axis