What is the minimum frequency of a photon that can eject a photoelectron from Ba metal? (The work function of barium is = 4.3 *10^-¹⁹ J.) A) 2.8 *10^-⁵²Hz B) 6.5 * 10¹⁴Hz C) 6.5 * 10⁴ Hz D) 6.5 * 10¹³ Hz E) 2.8 * 10¹⁴Hz

Question

Quizplus · Accepted Answer

The minimum frequency required to eject a photoelectron is given by the equation: 
f = (K.E. + Φ)/h 
where f is the frequency of the photon, K.E. is the maximum kinetic energy of the photoelectron, Φ is the work function, and h is Planck's constant. 
In this case, the work function is given as 4.3×10^-19 J. We know that the maximum kinetic energy of the photoelectron is equal to the energy of the photon minus the work function, i.e. K.E. = hv - Φ. We want to find the minimum frequency, so we need to use the maximum kinetic energy possible, which occurs when the photon has the minimum frequency required to eject a photoelectron. 
Substituting the values given: 
K.E. = 0 (since we want the minimum frequency which just about ejects a photoelectron, not one that gives it any extra kinetic energy) 
Φ = 4.3×10^-19 J 
h = 6.626×10^-34 J s 
We get: 
f(min) = Φ/h = 4.3×10^-19 J / 6.626×10^-34 J s ≈ 6.5×10^14 Hz 
Therefore, the best answer choice is B.

What Is the Minimum Frequency of a Photon That Can