If a metal has a work function of 2 eV, what is the minimum wavelength of light
Practice Questions
Q1
If a metal has a work function of 2 eV, what is the minimum wavelength of light required to cause the photoelectric effect?
620 nm
310 nm
1240 nm
500 nm
Questions & Step-by-Step Solutions
If a metal has a work function of 2 eV, what is the minimum wavelength of light required to cause the photoelectric effect?
Step 1: Understand the work function. The work function is the minimum energy needed to remove an electron from a metal. In this case, it is given as 2 eV.
Step 2: Convert the work function from electron volts (eV) to joules (J). Use the conversion factor: 1 eV = 1.6 x 10^-19 J. So, 2 eV = 2 * 1.6 x 10^-19 J.
Step 3: Calculate the energy in joules. Multiply: 2 * 1.6 x 10^-19 J = 3.2 x 10^-19 J.
Step 4: Use the formula for wavelength (λ). The formula is λ = hc/E, where h is Planck's constant (6.63 x 10^-34 J·s) and c is the speed of light (3 x 10^8 m/s).
Step 5: Substitute the values into the formula. λ = (6.63 x 10^-34 J·s * 3 x 10^8 m/s) / (3.2 x 10^-19 J).
Step 6: Calculate the wavelength. First, calculate the numerator: 6.63 x 10^-34 * 3 x 10^8 = 1.989 x 10^-25. Then divide by the energy: 1.989 x 10^-25 / 3.2 x 10^-19 = 6.21 x 10^-7 m.
Step 7: Convert the wavelength from meters to nanometers. 1 meter = 1 x 10^9 nanometers, so 6.21 x 10^-7 m = 621 nm.
Step 8: Verify the calculation. The minimum wavelength required to cause the photoelectric effect is approximately 310 nm.
Photoelectric Effect – The phenomenon where electrons are emitted from a material when it absorbs light of sufficient energy.
Work Function – The minimum energy required to remove an electron from the surface of a metal.
Energy-Wavelength Relationship – The relationship between the energy of a photon and its wavelength, given by the equation E = hc/λ.
