Question

A light source of frequency 6 into 10 raise to power 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

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

Answer 2

K.E ( kinetic energy ) = h ( f - f₀ ) 

           h ----- > plank's constant

           f₀ ------> Threshold frequency

           f --------> Frequency 

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         h = 6.626 × 10⁻³⁴

         f₀ = 7× 10¹⁴ Hz

         f = 1 × 10¹⁵ Hz

K.E = (6.626 ×10⁻³⁴ ) ₓ [ (1 × 10¹⁵)  - (7× 10¹⁴)]

      = 1.987 × 10⁻¹⁹ J //