Question
An ionized calcium atom has a charge of
+2 elementary charges. If this ion is accelerated
through a potential difference of 2.0 × 103 volts,
the ion’s change in kinetic energy will be
(1) 1.0 × 103 eV (3) 3.0 × 103 eV
(2) 2.0 × 103 eV (4) 4.0 × 103 eV
+2 elementary charges. If this ion is accelerated
through a potential difference of 2.0 × 103 volts,
the ion’s change in kinetic energy will be
(1) 1.0 × 103 eV (3) 3.0 × 103 eV
(2) 2.0 × 103 eV (4) 4.0 × 103 eV
Answers
GPT-4o mini
To find the change in kinetic energy of the ionized calcium atom, we can use the relationship between electric potential difference (voltage) and kinetic energy. The change in kinetic energy (\( \Delta KE \)) when a charge (\( q \)) is accelerated through a potential difference (\( V \)) is given by the formula:
\[
\Delta KE = qV
\]
In this case, the ionized calcium atom has a charge of +2 elementary charges. The elementary charge (\( e \)) is approximately \( 1.6 \times 10^{-19} \) coulombs. Therefore, the charge of the ionized calcium atom is:
\[
q = 2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ coulombs}
\]
The potential difference is given as:
\[
V = 2.0 \times 10^3 \text{ volts}
\]
Now, we can calculate the change in kinetic energy:
\[
\Delta KE = qV = (3.2 \times 10^{-19} \text{ C}) \times (2.0 \times 10^3 \text{ V}) = 6.4 \times 10^{-16} \text{ joules}
\]
To convert joules to electron volts (eV), we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ joules} \):
\[
\Delta KE \text{ in eV} = \frac{6.4 \times 10^{-16} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 4000 \text{ eV}
\]
So, the change in kinetic energy is:
\[
\Delta KE = 4.0 \times 10^3 \text{ eV}
\]
Thus, the correct answer is:
(4) \( 4.0 \times 10^3 \text{ eV} \)
\[
\Delta KE = qV
\]
In this case, the ionized calcium atom has a charge of +2 elementary charges. The elementary charge (\( e \)) is approximately \( 1.6 \times 10^{-19} \) coulombs. Therefore, the charge of the ionized calcium atom is:
\[
q = 2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \text{ coulombs}
\]
The potential difference is given as:
\[
V = 2.0 \times 10^3 \text{ volts}
\]
Now, we can calculate the change in kinetic energy:
\[
\Delta KE = qV = (3.2 \times 10^{-19} \text{ C}) \times (2.0 \times 10^3 \text{ V}) = 6.4 \times 10^{-16} \text{ joules}
\]
To convert joules to electron volts (eV), we use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ joules} \):
\[
\Delta KE \text{ in eV} = \frac{6.4 \times 10^{-16} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 4000 \text{ eV}
\]
So, the change in kinetic energy is:
\[
\Delta KE = 4.0 \times 10^3 \text{ eV}
\]
Thus, the correct answer is:
(4) \( 4.0 \times 10^3 \text{ eV} \)