A deuteron of kinetic energy 50 , keV is desc | vedclass

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(D) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $m$ is the mass,$K$ is the k

Vedclass · Accepted Answer

$100$

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(D) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{\sqrt{2mK}}{qB}$,where $m$ is the mass,$K$ is the kinetic energy,and $q$ is the charge.From this,we can write $K = \frac{r^2 q^2 B^2}{2m}$.Since $r$ and $B$ are the same for both particles,$K \propto \frac{q^2}{m}$.For a deuteron $(d)$,$q_d = e$ and $m_d = 2m_p$. For a proton $(p)$,$q_p = e$ and $m_p = m_p$.Taking the ratio: $\frac{K_p}{K_d} = \left( \frac{q_p}{q_d} ight)^2 	imes \left( \frac{m_d}{m_p} ight) = \left( \frac{e}{e} ight)^2 	imes \left( \frac{2m_p}{m_p} ight) = 1^2 	imes 2 = 2$.Therefore,$K_p = 2 	imes K_d = 2 	imes 50 \, keV = 100 \, keV$.

$A$ deuteron of kinetic energy $50 \, keV$ is describing a circular orbit of radius $0.5 \, m$ in a plane perpendicular to a magnetic field $\vec{B}$. The kinetic energy of a proton that describes a circular orbit of radius $0.5 \, m$ in the same plane with the same $\vec{B}$ is ........ $keV$.

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