Question

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate the charge on the ion.

Answer 1

Answer: $6.408(10)^{-19} C$

Explanation:

This problem can be solved by the following equation:

$\Delta K=q V$

Where:

$\Delta K=7.37(10)^{-17} J$ is the change in kinetic energy

$V=115 V$ is the electric potential difference

$q$ is the electric charge

Finding $q$:

$q=\frac{\Delta K}{V}$

$q=\frac{7.37(10)^{-17} J}{115 V}$

Finally:

$q=6.408(10)^{-19} C$

Answer 2

The charge on the ion will be "6.408 × 10⁻¹⁹ C".

Kinetic energy

Kinetic energy (K.E) would be a type of energy that being an item as well as particles possesses as a consequence among its motion. Whenever work, which transfers energy, has been performed on such an item by exerting a net force, the object accelerates, obtains K.E.

According to the question,

Increase in Kinetic energy, ΔK = 7.37 × 10⁻¹⁷ J

Electric potential, V = 115 V

We know the relation,

→ ΔK = qV

or,

Charge, q = $\frac{\Delta K}{V}$

By substituting the values, we get

                 = $\frac{7.37\times 10^{-17}}{115}$

                 = 6.408 × 10⁻¹⁹ C

Thus the above answer is appropriate.

Find out more information about Kinetic energy here:

https://brainly.com/question/114210