To calculate the speed of the ejected electron, we need to use the equation for the kinetic energy of an electron:
KE = hf - Φ
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 Φ is the work function of calcium.
To find the frequency of the light, we can use the equation:
c = λf
Where c is the speed of light (3.00 × 10^8 m/s) and λ is the wavelength of the light.
Let's calculate the frequency first. The wavelength is given as 285 nm, which is equivalent to 2.85 × 10^-7 m. Plugging this into the equation, we get:
c = λf
3.00 × 10^8 m/s = (2.85 × 10^-7 m)f
Now, solve for f:
f = (3.00 × 10^8 m/s) / (2.85 × 10^-7 m)
f ≈ 1.05 × 10^15 Hz
Now that we have the frequency, we can calculate the kinetic energy using the equation:
KE = hf - Φ
KE = (6.626 × 10^-34 J·s)(1.05 × 10^15 Hz) - (4.34 × 10^-19 J)
KE ≈ 6.96 × 10^-19 J
Finally, we can use the kinetic energy to find the speed of the ejected electron using the equation:
KE = (1/2)mv^2
Where m is the mass of the electron (9.10938356 × 10^-31 kg). Rearranging the equation, we get:
v = √((2KE) / m)
v = √((2(6.96 × 10^-19 J)) / (9.10938356 × 10^-31 kg))
v ≈ 4.29 × 10^6 m/s
So, the speed of the ejected electron is approximately 4.29 × 10^6 m/s.