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A photosensitive metallic surface has work function φ₀. If photons of energy 3φ₀ fall on this surface, the electrons come out with a maximum velocity of 6×10⁶ m s⁻¹. When the photon energy is increased to 9φ₀, the maximum velocity of the photoelectrons will be ______.

Asked in GUJCET 2011 · Speed and kinetic energy of the photoelectrons

Answer: (1) 12×10⁶ m s⁻¹

Step-by-step solution

Given: work function φ₀; photon energy 3φ₀ gives v₁ = 6×10⁶ m s⁻¹; new photon energy 9φ₀.

Idea: 1/2mvₘₐₓ² = E - φ₀, so vₘₐₓ ∝ √E - φ₀.

K₁ = 3φ₀ - φ₀ = 2φ₀ and K₂ = 9φ₀ - φ₀ = 8φ₀.

(v₂)/(v₁) = √(8φ₀)/(2φ₀) = √4 = 2

v₂ = 2×6×10⁶ = 12×10⁶ m s⁻¹

So the new maximum velocity is 12×10⁶ m s⁻¹.

Why the other options are wrong

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