Passage V

Fora photon to free an electron from a metal, the photon's frequency, f, must equal or exceed the threshold frequency, fT, for the metal. If f > fT, the photon has extra energy that is transferred to the electron as kinetic energy, resulting in the ejection of the electron from the metal. Kmax is the highest kinetic energy an ejected electron can have for a given f.

Table 1 displays fT for the metals calcium (Ca), magnesium (Mg), mercury (Hg), and platinum (Pt). Figure 1 is a graph of Kmax versus f for each of the 4 metals. Figure 2 is a graph of a photon's energy versus its f.

27. Based on Figure 1, for electrons ejected from Pt by photons having f = 28.0 x1014 Hz, Kmax would be:

Answer and Explanation

Your Answer is

Correct Answer is D

Explanation

Put value questions.

Method 1, clear at a glance: Look at the straight line corresponding to Pt in figure 1 is an upward trend, so the larger the abscissa, the larger the ordinate, and when the abscissa is 28, the ordinate must be greater than that in the figure The maximum time limit is 6

Method 2: Time-consuming and labor-intensive method: According to Figure 1, look at the two coordinate points (16,2) and (22,6) on the Pt straight line, and enter y= kx+b, calculate k=2/3, b=-26/3. So when the abscissa x=28, the ordinate=10