Question

question 2 of 15 view policies current attempt in progress calculate the kinetic energy of an electron moving at a speed of 2.26 × 10⁵ m/s (see the hint for electron properties).

Explanation:

Step1: Recall the formula for kinetic energy

The formula for the kinetic energy (\(K\)) of a particle is \(K = \frac{1}{2}mv^2\), where \(m\) is the mass of the particle and \(v\) is its velocity. For an electron, the mass \(m = 9.11\times 10^{-31}\space kg\) and the given velocity \(v=2.26\times 10^{5}\space m/s\).

Step2: Substitute the values into the formula

First, calculate \(v^{2}\):
\(v^{2}=(2.26\times 10^{5}\space m/s)^{2}=(2.26)^{2}\times(10^{5})^{2}=5.1076\times 10^{10}\space m^{2}/s^{2}\)

Then, substitute \(m = 9.11\times 10^{-31}\space kg\) and \(v^{2}=5.1076\times 10^{10}\space m^{2}/s^{2}\) into the kinetic energy formula:

\(K=\frac{1}{2}\times(9.11\times 10^{-31}\space kg)\times(5.1076\times 10^{10}\space m^{2}/s^{2})\)

First, multiply the constants: \(\frac{1}{2}\times9.11\times5.1076\approx\frac{1}{2}\times46.53\approx23.265\)

Then, multiply the powers of 10: \(10^{-31}\times 10^{10}=10^{-21}\)

So, \(K = 23.265\times 10^{-21}\space J=2.3265\times 10^{-20}\space J\) (we can also round it to appropriate significant figures. The given values: speed has 3 significant figures, mass of electron is known precisely, so we can present it as \(2.33\times 10^{-20}\space J\) or keep more decimals as per calculation)

Answer:

The kinetic energy of the electron is approximately \(\boldsymbol{2.33\times 10^{-20}\space J}\) (or more accurately from the calculation \(2.3265\times 10^{-20}\space J\))