Question

) Calculate the speed of electrons emitted from the surface of potassium metal, when light
of wavelength (a) 300 nm (b) 600 nm impinges a clean potassium metal surface? The
work function (9) of potassium is 2.3 eV.​

Answer 1

Explanation:

QUESTION :-

If threshold frequency of a metal is 1.11 × 10⁻⁶ Hz Then the maximum kinetic energy of photo electrons produced by a light of wavelength 15 A⁰ on the metal is

GIVEN :-

   Threshold frequency of a metal = 1.11 × 10⁶Hz

   Wavelength of light = 15A⁰

TO FIND :-

   Maximum Kinetic energy of the photo electron

SOLUTION :-

Where ,

   υₒ is threshold frequency

   λₒ is threshold wavelength

   c is velocity of light

We have ,

   υ₀ = 1.11 × 10⁶ s⁻¹

   c = 3 × 10⁻⁸ m/s

Relation between KE , λ , h , c and λ₀ is given by ,

Where,

   h is planck's constant

   c is velocity of light

   λ is wavelength

   λ₀ is threshold wavelength

We have,

   h = 6.625 × 10⁻³⁴ J.sec

   c = 3 × 10⁸ m/s

   λ = 15 × 10⁻¹⁰ m

   λ₀ = 270 m

∴ The Kinetic energy of the photoelectrons is 0.13 × 10⁻¹⁷ J