Question

If the maximum value of acceleration potential applied by a radio frequency oscillator be 20kV, find the number of revolutions made by a proton in cyclotron to achieve one fifth the speed of light Mass of a proton $=1.67\times 10^{-27}Kg$

• A. 420
• B. 470
• C. 400
• D. 200

Answer 1

Answer: (B)

$As=\frac{1}{2}m{v}^2=(2\times eV)n=2neV$
(as a proton gains an energy eV as it crosses fi-om one dee to the other and it does so twice in each revolution.)
$n=\frac{m{v}^2}{4eV}=\frac{(1.67\times 10^{-27})(0.6\times 10^8{)}^2}{4\times (1.6\times 10^{-19})(2\times 10^4)}=470$
$(asv=\frac{c}{5}=\frac{3\times 10^{8}m/s}{5}=0.6\times 10^{8}m/sand$
$V(=20KV=20\times 10^{3}V=2\times 10^{4}V)$