JEE MainPhysicsDual Nature of Matter
The graph showing the variation of stopping potential ( V_s ) with the frequency ( ) of incident light for a given photosensitive metal is a straight line intersecting the frequency axis at 4.0 10¹⁴ Hz . The work function of the metal is : (Take Planck's constant h = 6.6 10⁻³⁴ J s and elementary charge e = 1.6 10⁻¹⁹ C )
Options
- A1.65 eV
- B2.64 10⁻¹⁹ eV
- C4.22 10⁻³⁸ eV
- D4.125 10⁻¹⁵ eV
Correct answer
A. 1.65 eV
Step-by-step solution
The x-intercept of the stopping potential versus frequency graph represents the threshold frequency ( ₀ ) of the metal. Given ₀ = 4.0 10¹⁴ Hz . The work function ₀ in Joules is given by: ₀ = h ₀ ₀ = 6.6 10⁻³⁴ 4.0 10¹⁴ = 26.4 10⁻²⁰ J = 2.64 10⁻¹⁹ J To convert this into electron-volts (eV), divide by the elementary charge e : ₀ = 2.64 10⁻¹⁹ 1.6 10⁻¹⁹ eV ₀ = 1.65 eV Answer: 1.65 eV