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.)

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Xema8 Xema8 · Answer 1 · 2020-04-22T06:10:22+02:00

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 .