The electric field associated with a mono chromatic light wave is given by E=E₀ [ (1.57 10^7 ~m ⁻¹ ](x-c t) ] then the stopping potential when this light is used in a photoelectric experiment with the metal having work function 1.9 eV is. [Planck's constant, h=6.64 10⁻³⁴ ~J - s ]

Step-by-step solution

Given, E=E₀ [ (1.57 10^7 ~m ⁻¹ )(x-c t) ]...(i) Work function, ₀=1.9 eV =1.9 1.6 10⁻¹⁹ ~J Given, planck's constant , h=6.64 10⁻³⁴ ~J - s We know that, h v= ₀+e V₀ (Photoelectric equation)...(ii) Where, h v= incident energy of one photon ₀= work function of metal e= charge of an electronl V₀= stopping potential Now, from Eq. (i) Clearly 2 / =1.57 10^7 2 314 =1.57 10^7 = 2 314 1.57 10^7 =4 10⁻⁷ Now, from Eq. (ii) aligned h v & = ₀+e V₀ & h c = ₀+e V₀ ( v=c / ) aligned aligned & h c - ₀=e V₀ & 6.64 10⁻³⁴ 3 10^8 4 10⁻⁷