A deutron of kinetic energy 50 KeV is described a circular orbit 0.5 m in a plane perpendicular to a magnetic field B. What is the kinetic energy of a proton described a circluar orbit of radius 0.6 m in the same plane with the same B?

Options

A. `50 KeV`
B. `100 KeV`
C. `150 KeV`
D. `200 KeV`

Answer

B

Explanation

For a charged particle, describing a circular path in a transverse magnetic field,`(mv^(2))/r=qvB thereforev=(qvB)/m therefore Its K.E=1.2mv^(2)=1/2m.(q^(2)B^(2)r^(2))/(m^(2))=1/2(B^(2)q^(2)r^(2))/mtherefore`For a deutron, `mass(m_2)=2m_p` and both have the same B and r.`therefore E_("proton")=1/2(B^(2)q^(2)r^(2))/m_p and E_("deutron")=1/2(B^(2)q^(2)r^(2))/m_d therefore(E_p)/E_d=m_d/m_p=(2m_p)/m_p=2/1[thereforem_d=2m_p]therefore E_p=2E_d=2 xx 50 =100 KeV`