Energy of the incident photons on the metal surface is initially 4 W and then 6 W where W is the work function of that metal. The ratio of velocities of emitted photoelectrons is

Step-by-step solution

The kinetic energy of photoelectrons is given by K_ = 1 2 mv² for incident energy E exceeding the work function W , with K_ = E - W . For incident energy E₁ = 4W , the kinetic energy is K_ 1 = 4W - W = 3W , so 1 2 mv₁² = 3W . For incident energy E₂ = 6W , the kinetic energy is K_ 2 = 6W - W = 5W , so 1 2 mv₂² = 5W . Taking the ratio of the two kinetic energy equations yields 1 2 mv₁² 1 2 mv₂² = 3W 5W , which simplifies to v₁² v₂² = 3 5 . The velocity ratio is therefore v₁ v₂ = 3 5 = 3 5 . Thus, the ratio of the vel