If K 1 and K 2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength λ 1 and λ 2 respectively incident on a metallic surface. If λ 1 = 3 λ 2 , then

Question

If K 1 and K 2 are maximum kinetic energies of photoelectrons emitted when lights of wavelength λ 1 and λ 2 respectively incident on a metallic surface. If λ 1 = 3 λ 2 , then, K 1 > K 2 3, K 1 K 2 3, K 1 = 3 K 2, K 2 = 3 K 1

MARKS App · Accepted Answer

Eienstein's photoelectric equation is given as, K 1 = h c 1 λ 1 - 1 λ 0 ...(i) and K 2 = h c 1 λ 2 - 1 λ 0 ...(ii) Dividing Equation (i) by Equation (ii), we get K 1 K 2 = 1 λ 1 - 1 λ 0 1 λ 2 - 1 λ 0 Since, λ 1 = 3 λ 2 K 1 K 2 = 1 3 λ 2 - 1 λ 0 1 λ 2 - 1 λ 0 = 1 3 λ 2 1 - 3 λ 0 / λ 2 1 λ 2 1 - 1 λ 0 K 1 K 2 = 1 3 1 - 3 λ 0 / λ 2 1 - 1 λ 0 / λ 2 3 K 1 K 2 = 1 - 3 / λ 0 / λ 2 1 - 1 / λ 0 λ 2 1 ∴ 3 K 1 K 2 1 ⇒ K 1 K 2 3

Search any question & find its solution

Looking for more such questions to practice?