Question

A completely ionized beryllium atom (net charge = +4e) is accelerated through a potential difference of 6.0 v. what is the increase in kinetic energy of the atom?

Answer 1

The increase in kinetic energy of a completely ionized beryllium atom accelerated through a potential difference of 6.0 V is +24 electron volts (eV).

Step-by-step explanation:

To calculate the increase in kinetic energy of a completely ionized beryllium atom accelerated through a potential difference of 6.0 V, we need to use the formula:

ΔK = q × ΔV

Where ΔK is the increase in kinetic energy, q is the charge of the ion, and ΔV is the potential difference. In this case, the beryllium atom is completely ionized with a net charge of +4e, where e is the elementary charge. Therefore, the charge (q) is +4e.

Substituting the values, ΔK = (+4e) × (6.0 V) = +24eV

So, the increase in kinetic energy of the atom is +24 electron volts (eV).

Answer 2

For the principle of conservation of energy, the increase in kinetic energy of the atom is equal to the energy lost when moving through the potential difference:

`\Delta K=\Delta U=q \Delta V`
where q is the charge of the Beryllium atom:

`q=4 e=4 \cdot (1.6 \cdot 10^(-19)C)=6.4 \cdot 10^(-19)C`
and
`\Delta V=6.0 V`. Therefore, the increase in kinetic energy is

`\Delta K = (6.4 \cdot 10^(-19)C)(6.0 V)=3.8\cdot 10^(-18) J`