Question

In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.00 × 10−14 m from a fragment that has 46 protons in it? (b) What is the potential energy in MeV of a similarly charged fragment at this distance?

Answer 1

To find the electric potential near a fission fragment with 46 protons, we use the formula for the potential from a point charge. After calculating the charge of the fragment, we can find the potential and then the potential energy. To get the potential energy in MeV, we convert from joules to MeV.

Step-by-step explanation:

In nuclear fission, the nucleus of an atom splits into two or more smaller nuclei, along with other particles. This process can release a significant amount of energy. To calculate the electric potential at a distance from a charged fragment, we use the formula for the electrical potential due to a point charge:

V = k * Q / r

Where:

• V is the electric potential.
• k is Coulomb's constant (8.988 × 10^9 N·m^2/C^2).
• Q is the charge of the fragment.
• r is the distance from the charge, where the potential is being calculated.

The charge Q of a fragment with 46 protons can be calculated as Q = (number of protons) * (elementary charge), which is 46 * 1.602 × 10^-19 coulombs.

We then insert r, Q, and k into the equation to find the electric potential V at a distance of 2.00 × 10^-14 meters.

For part (b), the electric potential energy U can be calculated using the formula:

U = Q * V

Where:

• U is the potential energy.
• V is the potential computed in part (a).

We then convert the potential energy from joules to megaelectronvolts (MeV) using the conversion factor 1 eV = 1.602 × 10^-19 joules.

Answer 2

electric potential is 3.31 ×
`10^(6)` V

potential energy is 152 MeV

Step-by-step explanation:

given data

fragment charge Q = 46 protons = 46 × 1.6 ×
`10^(-19)` C

to find out

electric potential and potential energy

solution

we know here distance from fragment d = 2 ×
`10^(-14)` m

and constant for electric force k that is 9 ×
`10^(9)` N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 ×
`10^(9)[/tex ×46 × 1.6 × [tex]10^(-19)` / ( 2 ×
`10^(-14)` )

electric potential = 3.31 ×
`10^(6)` V

and

we know relation between electric potential and potential

that is V = U/q

so U will be = qV

now put all value

we get potential energy U

potential energy = 46 × 3.31 ×
`10^(6)`

potential energy = 1.52 ×
`10^(8)` eV

so potential energy = 152 MeV