If the maximum value of accelerating potential provided by a radio frequency oscillator be `10kV`, calculate the number of revolutions made by an `alpha`-particle in a cyclotron to achieve one-tenth of the speed of light. Mass of proton`=1*67xx10^(-27)kg`, charge on proton `=1*6xx10^(-19)C`.

Answer

470 revolutions

Explanation

Here, `V=10kV=10xx10^3V=10^4V`,
Mass of `alpha`-particle,
`m_alpha=4m_p=4xx1*67xx10^(-27)kg`,
Charge of `alpha`-particle,
`q_alpha=2q_p=2xx1*6xx10^(-19)C`.
In a cyclotron, an `alpha`-particle gains energy `q_alphaV` when it crosses a region of potential difference V. In one revolution, the `alpha`-particle crosses the gap twice between dees, so the energy gained in each revolution `=2q_alphaV`
Let `alpha`-particles makes n revolutions before coming out of the dees. The gain in its kinetic energy will be
`1/2m_alphav_alpha^2=n2q_alphaV` or `n=(m_alphav_alpha^2)/(4q_alphaV)`.
Given, `v_alpha=c/10=(3xx10^8)/(10)=3xx10^7ms^-1`
`n=((4xx1*67xx10^(-27))xx(3xx10^7)^2)/(4xx(2xx1*6xx10^(-19))xx10^4)`
=470 revolutions