A proton of energy E is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be equal to

Options

A. `E`
B. `2 E`
C. `E/2`
D. `0.75 E`

Answer

A

Explanation

`E=(B^(2)q^(2)r^(2))/(2m)` For an `alpha` paarticle, `q=2e and m_alpha=4m_p therefore E_alpha/E_p=(B^(2)(2e)^(2))/(2(4m_p))xx (2(mp))/(B^(2)(R)^(2)x r^(2))=4/4=1 therefore E_alpha=R_p`.