Calculate the kinetic energy of the photoelectron emitted by a metal (threshold energy is 1.50 x 10^-19J) exposed to light of wavelength 5.00 x 10^-7 m (the energy beyond that required to knock electron goes into electron kinetic energy).
A) -3/98 x 10^-19J
B) 3.98 x 10^-19J
C) 2.48 x 10^-19J
D) 1.50 x 10^-19J
E) 8.3141J
dE = hc/lambda = ?
KE = dE-threshold
To solve this problem, we need to use the equation relating the kinetic energy of an electron to the energy of a photon and the threshold energy. The equation is given by:
Kinetic energy = Energy of photon - Threshold energy
Given:
Threshold energy = 1.50 x 10^-19 J
Wavelength of light = 5.00 x 10^-7 m
First, we need to convert the wavelength of light into energy using the equation:
Energy of photon = (Planck's constant * speed of light) / wavelength
Planck's constant (h) ≈ 6.626 x 10^-34 J·s
Speed of light (c) ≈ 3.0 x 10^8 m/s
Substituting the values into the equation:
Energy of photon = (6.626 x 10^-34 J·s * 3.0 x 10^8 m/s) / (5.00 x 10^-7 m)
Calculating the value:
Energy of photon ≈ 3.9768 x 10^-19 J
Now, substitute the values into the equation for kinetic energy:
Kinetic energy = 3.9768 x 10^-19 J - 1.50 x 10^-19 J
Calculating the value:
Kinetic energy ≈ 2.4768 x 10^-19 J
Therefore, the kinetic energy of the photoelectron is approximately 2.48 x 10^-19 J.
So, the correct answer is option C: 2.48 x 10^-19 J.