Question

Express in $eV$(or $keV$or $MeV$if more appropriate):

a. The kinetic energy of a Li ++ ion that has accelerated from rest through a potential difference of $5000V$.

b. The potential energy of two protons $10fm$apart.

c. The kinetic energy, just before impact, of a $200g$ ball dropped from a height of $1.0m$.

Short answer

a. The potential energy of two protons is$10KeV$.

b. The potential energy of two protons is$14.4KeV$.

c. The kinetic energy just before impact is$1.23\times {10}^{19}eV$.

Step-by-step solution

Part (a) step 1: Given information

We have given that express in $eV$

We need to find the kinetic energy of a Li ++ ion that has accelerated from rest through a potential difference of $5000V$

Part (a) step 2: Simplify

$L{i}^{++}$Ion has $+2e$charge. As it is accelerated through the $5000V$the total energy will be

$2\times 5000=10000eV=10keV$

Part (b) step 2: Given information

We have given that express in $eV$$eV$

We need to find the potential energy of two protons$10fm$ apart.

Part (b) step 2: Simplify

Potential energy is given as

$$\begin{gathered} V=\frac{q_1{q}_2}{4\mathrm{\pi }{\in }_0\mathrm{r}} \\ =\frac{1.6\times {10}^{-19}\times 1.6\times {10}^{-19}}{4\mathrm{\pi }\times 8.85\times {10}^{-12}\times 10\times {10}^{-15}} \\ =2.3\times {10}^{-14}\mathrm{J} \\ =\frac{2.30\times {10}^{-14}}{1.6\times {10}^{-19}}=1.44\times {10}^5\mathrm{eV} \\ =14.4\mathrm{keV} \end{gathered}$$

Part (c) step 1: Given information

We have given that express in $eV$

We need to find the kinetic energy, just before impact, of a $200g$ball dropped from a height of n$1.0m$

Part (c) step 2: Simplify

The kinetic energy just before hitting the ground will be equal to the potential energy and is equal to $mgh$. Here $m=200g=0.2kg,g=9.8m{s}^{-2}$and $h=1.0m$.So energy will be

$$\begin{gathered} E=mgh \\ =0.2\times 9.8\times 1=1.96J \\ =\frac{1.96}{1.6\times {10}^{-19}}=1.23\times {10}^{19}eV \end{gathered}$$