To determine the change in kinetic energy of a proton when it is accelerated through a potential difference of 2 megavolts (MV), we can use the basic physics equation for kinetic energy change due to electric potential difference:
[tex]\[ \Delta KE = q \times V \][/tex]
where:
- [tex]\(\Delta KE\)[/tex] is the change in kinetic energy,
- [tex]\(q\)[/tex] is the charge of the proton,
- [tex]\(V\)[/tex] is the potential difference.
Given:
- The charge of a proton ([tex]\(q\)[/tex]) is [tex]\(1.6 \times 10^{-19}\)[/tex] Coulombs.
- The potential difference ([tex]\(V\)[/tex]) is [tex]\(2 \times 10^{6}\)[/tex] volts (2 MV).
Let's plug these values into the equation:
[tex]\[ \Delta KE = (1.6 \times 10^{-19} \, \text{C}) \times (2 \times 10^{6} \, \text{V}) \][/tex]
First, multiply the numerical parts:
[tex]\[ 1.6 \times 2 = 3.2 \][/tex]
Next, combine the powers of 10:
[tex]\[ 10^{-19} \times 10^{6} = 10^{-19 + 6} = 10^{-13} \][/tex]
So, the change in kinetic energy is:
[tex]\[ \Delta KE = 3.2 \times 10^{-13} \, \text{Joules} \][/tex]
To convert this energy into nanojoules (nJ), we use the conversion factor [tex]\(1 \, \text{Joule} = 10^{9} \, \text{nanojoules}\)[/tex]:
[tex]\[ \Delta KE = 3.2 \times 10^{-13} \, \text{J} \times 10^{9} \, \left(\frac{\text{nJ}}{\text{J}}\right) = 3.2 \times 10^{-13 + 9} \, \text{nJ} = 3.2 \times 10^{-4} \, \text{nJ} = 0.00032 \, \text{nJ} \][/tex]
Therefore, the change in kinetic energy of the proton, when accelerated through a potential difference of 2 MV, is:
[tex]\[ \Delta KE = 3.2 \times 10^{-13} \, \text{J} \][/tex]
or equivalently,
[tex]\[ \Delta KE = 0.00032 \, \text{nJ} \][/tex]
Hence, when considering the decimal notation, the correct numerical result in nanojoules (nJ) is:
[tex]\[ \boxed{0.00032 \, \text{nJ}} \][/tex]
