A photosensitive metal has a threshold frequency ₀ . When it is illuminated by light of frequency 2 ₀ , the stopping potential is V₁ . When the same metal is illuminated by light of an unknown frequency _x , the stopping potential becomes exactly double ( 2V₁ ). The unknown incident frequency _x is:

Step-by-step solution

Using Einstein's photoelectric equation: eV_s = h - h ₀ For the first case, the incident frequency is 2 ₀ and the stopping potential is V₁ : eV₁ = h(2 ₀) - h ₀ eV₁ = h ₀ ... (i) For the second case, the incident frequency is _x and the stopping potential is 2V₁ : e(2V₁) = h _x - h ₀ 2(eV₁) = h _x - h ₀ ... (ii) Substituting the value of eV₁ from equation (i) into equation (ii): 2(h ₀) = h _x - h ₀ 2h ₀ + h ₀ = h _x 3h ₀ = h _x _x = 3 ₀ Answer: 3 ₀