91Ó°ÊÓ

We need to determine the speed of a $300eV$electron.

Kinetic energy is $300eV=300×1.6×{10}^{-19}J$. So the velocity is given as

$v=\sqrt{\frac{2E_k}{m}}=\sqrt{\frac{2×300×1.6×{10}^{-19}}{9.1×{10}^{-31}}}=1.03×{10}^{7}m/s$

We need to determine the speed of a $3.5MeV{H}^{+}$ion.

The kinetic energy is $3.5MeV=3.5×{10}^6×1.6×{10}^{-19}J$, and the mass of $H^{+}$is $m_p=1.67×{10}^{-27}$, so the velocity is given as

$v=\sqrt{\frac{2E_k}{m_p}}=\sqrt{\frac{2×3.5×1.6×{10}^{-13}}{1.67×{10}^{-27}}}=2.59×{10}^{7}m/s$

We need to determine the specific type of particle that has $2.09MeV$of kinetic energy when moving with a speed of $1.0×{10}^{7}m/s$.

Here the kinetic energy is $2.09MeV=2.09×{10}^6×1.6×{10}^{-19}J$, the velocity is $v=1.0×{10}^{7}m/s$. So the mass of the particle will be

$m=\frac{2E_k}{v^2}=\frac{2×2.09×1.6×{10}^{-13}}{{(1.0×{10}^7)}^2}=6.68×{10}^{-27}kg$

As the mass of the particle matches with the alpha particle. So, the particle is an alpha particle.$2.09MeV=2.09×{10}^6×1.6×{10}^{-19}J$