In a photoelectric experiment light of wavelength 800 nm produces photoelectrons with the smallest de Broglie wavelength of 1 nm . Then the work function of the metal used in the experiment is nearly

Step-by-step solution

Kinetic energy of the electron having de Broglie wavelength λ can be written as, λ b = h p = h 2 m K ⇒ K = h 2 2 m λ b 2 For the given values, we can write: h c λ = K + ϕ ⇒ h c 800 × 10 - 9 = h 2 2 m × 1 × 10 - 9 2 + ϕ ⇒ ϕ = h c 800 × 10 - 9 - h 2 2 m × 1 × 10 - 9 2 = 6 . 6 × 10 - 34 × 3 × 10 8 800 × 10 - 9 - 6 . 6 × 10 - 34 2 2 9 . 1 × 10 - 31 × 1 × 10 - 9 2 = 2 . 47 × 10 - 19 - 2 . 39 &#