Practice portal › Dual Nature of Matter and Radiation › Einstein's Photoelectric Equation and Work Function

When light of frequencies f₁ and f₂ falls on two identical photosensitive surfaces, the maximum velocities of the photoelectrons (mass m) are v₁ and v₂. Hence ______

Asked in RS Academy GUJCET booklet · Speed and kinetic energy of the photoelectrons

• (1) v₁²-v₂²=(2h)/m(f₁-f₂)
• (2) v₁+v₂=[(2h)/m(f₁-f₂)]^1/2
• (3) v₁²-v₂²=(2h)/m(f₁+f₂)
• (4) v₁-v₂=[(2h)/m(f₁-f₂)]^1/2

Step-by-step solution

Why the other options are wrong

• (2) Taking the square root of the correct relation gives √v₁²-v₂², which is not v₁ + v₂. With v₁ = 5 and v₂ = 3: √25-9 = 4, but v₁ + v₂ = 8.
• (3) Subtracting the two equations cancels φ₀ and leaves the difference h(f₁ - f₂). A plus sign would need adding them, but adding keeps 2φ₀ and puts v₁² + v₂² on the left.
• (4) The same root slip with the other factor: √v₁²-v₂² is not v₁ - v₂. With v₁ = 5 and v₂ = 3: √25-9 = 4, but v₁ - v₂ = 2.

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.