Answer to Question #204466 in General Chemistry for lea

Question

please show the steps. formula:

12. Selenium was one of the first elements used in the construction of “electric eyes”, due to its photoconductive properties. The work function for selenium is 5.11 eV. What is the longest wavelength of light that will just cause electrons to be ejected from a surface coated with selenium? (Formally, the electrons leave with no kinetic energy.)

a. In what range of the electromagnetic spectrum does the light of the wavelength from part a fall?

b. When a particular monochromatic light source shines on a selenium surface, photoelectrons are ejected with a kinetic energy of 1.05 eV. What is the wavelength of the incident monochromatic light?

c. Calculate the maximum velocity (m/s) of the ejected electrons from the previous question.

Answer 1

1.(a) 1 eV = 1.6 × 10^-19 J

5.11 eV × (1.6 × 10^-19 J / 1eV) = 8.176 ×10^-19 J

Now, energy (E) = hc/ \lambda

Where h = Planck's constant = 6.626 ×10^-34 J.s

c = velocity of light = 3.0 ×10^8 m/s

Now, 8.176 ×10^-19 J = ( 6.626 ×10^-34 J.s × 3.0 ×10^8 m/s) / \lambda

\lambda = 243.1 nm = wavelength

b)It lies in UV = ultraviolet region

(c) K.E of electron = 1.05 eV =1.05 eV *(1.6 * 10-19 J / 1eV) = 1.68 ×10^-19 J

Energy gained by electron to eject = work function = 8.176 ×10^-19 J

Total energy of light incident = 8.176 ×10^-19 J + 1.68 ×10^-19 J = 9.856×10^-19 J

energy (E) = hc/ \lambda

9.856×10^-19 J = (6.626 ×10^-34 J.s × 3.0 ×10^8 m/s) / \lambda

\lambda = 201.7 nm = wavelength of incident light