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  4. 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 × 10-27Kg

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Q. 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 $=1.67 \times 10^{-27}Kg$

Moving Charges and Magnetism

Solution:

$As=\frac{1}{2}mv^2=(2\times eV)n=2neV$
(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.)
$n=\frac{mv^2}{4eV}=\frac{(1.67 \times 10^{-27})(0.6 \times 10^8)^2}{4 \times (1.6 \times 10^{-19})(2 \times 10^4)}=470$
$(as \,v=\frac{c}{5}=\frac{3 \times 10^8\,m/s}{5}=0.6 \times 10^8\,m/s\,and$
$V(=20KV=20 \times 10^3V=2\times 10^4V) $