Question

what is the average kinetic energy in joules (j) of an electron (mass 9.11x10^-31 kg) that has a velocity of 4.2 x 10^6 m/s? formula bank: ke = 1/2·mv² (m = mass, v = velocity m/s) l = h/(mv) l= lamda wavelength nm, h = plancks constant = 6.626x10^34 j·s 6.4 x10^-17 3.2 x10^-17 6.4 x10^-16 8.0 10^-18 1.6 x10^-17

Explanation:

Step1: Substitute the values into the formula

Given \(m = 9.11\times10^{-31}\text{ kg}\) and \(v = 4.2\times 10^{6}\text{ m/s}\), and the formula \(KE=\frac{1}{2}mv^{2}\).
Substitute the values: \(KE=\frac{1}{2}\times(9.11\times 10^{-31})\times(4.2\times 10^{6})^{2}\)

Step2: Calculate \((4.2\times 10^{6})^{2}\)

Using the rule \((a\times10^{n})^{2}=a^{2}\times 10^{2n}\), we have \((4.2\times 10^{6})^{2}=4.2^{2}\times10^{12}=17.64\times 10^{12}\)

Step3: Calculate the product

\(KE=\frac{1}{2}\times9.11\times10^{-31}\times17.64\times 10^{12}\)
First, calculate \(\frac{1}{2}\times9.11\times17.64 = 80.0\) (approximate value). Then, using the rule \(a\times10^{m}\times b\times10^{n}=ab\times10^{m + n}\), we have \(80.0\times10^{-31 + 12}=80.0\times10^{-19}=8.0\times10^{-18}\) (There was a miscalculation in the previous step - let's re - calculate properly.

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Answer:

\(8.0\times 10^{-18}\) (corresponding to the option \(8.0\times10^{-18}\))