When light of wavelength 236 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.99 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface?
(Use 1 eV = 1.602 ✕ 10−19 J, e = 1.602 ✕ 10−19 C, c = 2.998 ✕ 108 m/s, and h = 6.626 ✕ 10−34 J · s = 4.136 ✕ 10−15 eV · s as necessary.)
Question
Answer:
Explanation:
Energy of photon = h c / λ , h is planks constant , c is velocity of light and λ is wave length
= 6.626 x 10⁻³⁴ x 2.998 x 10⁸ / 236 x 10⁻⁹
= .08417 x 10⁻¹⁷ J
Kinetic energy of photoelectron = 1.99 e V
= 1.99 x 1.602 x 10⁻¹⁹ J
= 3.18798 x 10⁻¹⁹
= .03188 x 10⁻¹⁷ J
Diff of energy = .08417 – .03188 x 10⁻¹⁷
= .05229 x 10⁻¹⁷ J
This will be the work function of the metal
If λ be the maximum wave-length required
h c / λ = .05229 x 10⁻¹⁷
6.626 x 10⁻³⁴ x 2.998 x 10⁸ / λ = .05229 x 10⁻¹⁷
λ = 6.626 x 10⁻³⁴ x 2.998 x 10⁸ / .05229 x 10⁻¹⁷
= 379.89 x 10⁻⁹ m
= 379.89 nm .