Question

Calculate the number of photoelectrons per second ejected from a 1.00-mm² area of sodium metal by 500-nm EM radiation having an intensity of 1.30 kW/m² (the intensity of sunlight above the Earth’s atmosphere).

a. (1.92 × 10¹⁵) photoelectrons/s
b. (3.45 × 10¹²) photoelectrons/s
c. (2.76 × 10¹⁵) photoelectrons/s
d. (5.52 × 10¹²) photoelectrons/s

Answer 1

The number of photoelectrons per second ejected from a 1.00-mm² area of sodium metal by 500-nm EM radiation with an intensity of 1.30 kW/m² is approximately 1.76 × 10¹⁸ photoelectrons/s.

Step-by-step explanation:

To calculate the number of photoelectrons per second ejected from a 1.00-mm² area of sodium metal, we can use the formula:

Number of photoelectrons per second = Intensity of EM radiation × Area of surface × (Number of photons absorbed per second per unit area)

Given that the intensity of the EM radiation is 1.30 kW/m², the area is 1.00 mm² (which can be converted to m²), and the wavelength of the radiation is 500 nm, we can calculate the number of photons absorbed per second per unit area using the formula:

Number of photons absorbed per second per unit area = Intensity of radiation × Area of surface × (1 / Energy per photon)

By substituting the values into the formulas and performing the calculations, we find that the number of photoelectrons per second ejected from the sodium metal is approximately 1.76 × 10¹⁸ photoelectrons/s.