Question

a light source of frequency 6 into 10 to the power minus 15 hertz is subjected to a metal sheet having work function equal to 22.6 electron volt then the velocity of photoelectrons emitted having maximum kinetic energy will be

Answer 1

here is your answer seee.....

Given,

frequency , ν = 6 × 10¹⁵ Hz

work function , Ф = 22.6 eV

According to photoelectric concepts,

Kinetic energy of photons = energy of incident photons - work function.

e.g., 1/2 mv² = hν - Φ

Here, m = 9.1 × 10⁻³¹ Kg , h = 6.636 × 10⁻³⁴ J.s

so, 1/2 × 9.1 × 10⁻³¹ × v² = 6.636 × 10⁻³⁴ × 6 × 10¹⁵ - 22.6 × 1.6 × 10⁻¹⁹

[ We know, 1eV = 1.6 × 10⁻¹⁹ J]

4.55 × 10⁻³¹v² = 39.816 × 10⁻¹⁹ - 36.16 × 10⁻¹⁹

v² = 2.656/4.55 × 10¹² = 0.5837 × 10¹²

v = 7.64 × 10⁵ m/s