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------------------------------------------------ What is the speed of the helium ion (Q +2e) accelerated by a voltage of 3700 V?

A. \(5.78 \times 10^6\) m/s
B. \(7.24 \times 10^5\) m/s
C. \(4.51 \times 10^7\) m/s
D. \(3.91 \times 10^4\) m/s

None of the options is correct. The correct speed of the helium ion accelerated by a voltage of 3700 V is approximately [tex]\(3.77 \times 10^6 \, \text{m/s}\).[/tex]

To find the speed of the helium ion, we can use the formula for the kinetic energy of a charged particle accelerated by a voltage. The kinetic energy (K) can be calculated using the formula (K = q × V), where \(q\) is the charge of the ion and (V) is the voltage.

Since the helium ion has a charge of +2e and the voltage is 3700 V, we can substitute these values into the formula to find the kinetic energy. Next, we can use the formula for kinetic energy to find the speed (v) of the ion: [tex]\(K = \frac{1}{2} \times m \times v^2\)[/tex], where (m) is the mass of the ion and (v) is its speed.

By rearranging the formula and solving for (v), we can determine the speed of the ion. After performing the calculations, we find that the speed of the helium ion accelerated by a voltage of 3700 V is approximately [tex]\(3.77 \times 10^6 \, \text{m/s}\)[/tex].