91Ó°ÊÓ

reflection telescope = $20cm$

$focallength=1.00m$

using with a $10x$ eyepiece we need to find magnification

Since the energy is conserved, all the energy due to the potential drop will be converted to kinetic energy

This means that

$q∆V=\frac{m{v}^2}{2}$

from this, we can find the speed of the particles to be

$\mathrm{ν}=\sqrt{\frac{2q∆V}{m}}$

The absolute value of the charges will be the same, $e$. Also, the product $e∆V$will not change the sign since to accelerate two opposite charges we will have to apply opposite polarity accelerating potential differences. having said that we will numerically have

for the case of the portion

$\mathrm{ν}=\sqrt{\frac{2.1.6.{10}^{-19}.9}{.1.67.{10}^{-27}}}=4.2.{10}^{4}m/s$

reflection telescope = $20cm$

$focallength=1.00m$

using with a$10X$eyepiece we need to find $f-number$of the telescope

For the case of the electrons

$\mathrm{ν}=\sqrt{\frac{2.1.6.{10}^{-19}.9}{9.1.{10}^{-31}}}=1.8.{10}^{6}m/s$