Question

Light of wavelength λ falls on a metal with work-function hc/λ0. Photoelectric effect will take place only if
(a) λ ≥ λ0
(b) λ ≥ 2λ0
(c) λ ≤ λ0
(d) λ < λ0/2

Answer 1

• We know, photoelectric effect will take place only when the energy of the incident light is greater than or equal to the work function of the metal.

• Method 1: By Einstein's equation if photoelectric effect,                                   hc/λ = hc/λ₀ +KE

        This gives, on solving a bit,

        hc(λ₀ - λ)/λλ₀ = KE

        We know, KE or kinetic energy ≥ 0 (equal to 0 if velocity is 0)

        So, clearly λ₀ ≥ λ

• Method 2: We know for photoelectric effect to take place, energy of incident light must be greater than or equal to work function of the metal. So clearly, for energy of incident light to be more than work function of the metal, wavelength of incident light must be smaller than or equal to wavelength corresponding to work function, because energy is inversely proportional to wavelength.

Answer is option (c) λ ≤ λ₀

Answer 2

Light of wavelength λ falls on a metal with work-function hc/λ0. Photoelectric effect will take place only if $\lambda \leq \lambda 0$

Explanation:

• The threshold wavelength of the metal is λ0 as it’s work-function is hc/λ0.
• The incident light’s wavelength must be equal to or less than the metal’s threshold wavelength for the photoelectric effect.
• It is to be observed that the electron emission or the emission of other free carriers is the photoelectric effect. In this manner, the emission of electrons is called the photoelectrons.