Recent content by Variable_X

Right so a) is 248keV. 3.97×10−14JT=3.97×10−14J. b) v = √(2K/m) should be √(2(3.97x10^-14)/1.67×10-31) = 6.89528x10^9 ms^-1 I'm assuming the question has typos and it's meant to be for the speed of the PROTON, not the PHOTON. c) r = mv/(eB) r=1.67×10-31x6.9x10^9/(1.67×10-19x1.5) = 0.0046 nm

3.00x10^8 ms-1

Yeah, I understand now, it's 6.9x10^20ms-1 (rouned)

I not changing eV to Joules at that point I'm changing kV to eV

a) 248*10^3 eV for 248kV Calculate the energy in J K=248*10^3*1.6*10^-19 =396.8*10^-19 J b) K=(1/2)mv^2 v=sqrt(2k/m) =sqrt((2*396.8*10^-19)/1.67*10^-27) =218^10^3 m/s c) r=mv/qB =1.67*10^-27*218*10^3/1.6*10^-19*1.5*10^-4 =15.17 mr=mv/qB...