Question

The work function of potassium is 3.68 × 10−19 J. Calculate the kinetic energy of the ejected electrons when light of frequency equal to 8.62 × 1014 s−1 is used for irradiation.

Answer 1

The kinetic energy of the ejected electrons is 2.029 × 10^-19 J.

Step-by-step explanation:

The kinetic energy (KE) of the ejected electrons can be calculated using the equation KE = hf - W, where KE is the kinetic energy, h is Planck's constant (6.626 × 10^-34 J·s), f is the frequency of the light, and W is the work function of potassium (3.68 × 10^-19 J).

First, we need to convert the frequency given (8.62 × 10^14 s^-1) to energy using the Equation:

E = hf. So, E = (6.626 × 10^-34 J·s) * (8.62 × 10^14 s^-1) = 5.709 × 10^-19 J.

Substituting the values into the kinetic energy equation, we have:

KE = (5.709 × 10^-19 J) - (3.68 × 10^-19 J) = 2.029 × 10^-19 J.

Therefore, the kinetic energy of the ejected electrons is 2.029 × 10^-19 J.