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  • A proton of mass m and charge +e is moving in a circular orbit a magnetic field with energy <img alt="1 \mathrm{MeV}" src="https://entrancecorner.oncodecogs.com/gif.latex?1%20%5Cmathrm%7B

A proton of mass m and charge +e is moving in a circular orbit a magnetic field with energy 1 \mathrm{MeV} . What should be the energy of an \alpha-particle (mass 4 m and charge +2 e ) so that it revolves in a circular orbit of the same radius in the same magnetic field?

Option: 1

1 \mathrm{MeV}


Option: 2

2 \mathrm{MeV}


Option: 3

4 \mathrm{MeV}


Option: 4

0.5 \mathrm{MeV}


Answers (1)

best_answer

For proton:  \mathrm{r=\frac{m v}{e B}}
For  \mathrm{\alpha-particle ~r^{\prime}=\frac{m^{\prime} v^{\prime}}{e^{\prime} B}=\frac{4 m v^{\prime}}{2 e B}=\frac{2 m v^{\prime}}{e B}}
Given \mathrm{ r=r^{\prime}} . Hence \mathrm{ v^{\prime}=\frac{v}{2}} 
Energy of proton  \mathrm{ E=\frac{1}{2} m v^2} . Energy of \mathrm{ \alpha}-particle is

\mathrm{ E^{\prime}=\frac{1}{2} m^{\prime} v^{\prime 2}=\frac{1}{2} \times 4 m \times\left(\frac{v}{2}\right)^2=\frac{1}{2} m v^2=E}

Hence \mathrm{E^{\prime}=1 \mathrm{MeV}}  which is choice (a).

Posted by

Pankaj Sanodiya

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