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Step-by-Step Solution
Step 1: Identify the frequencies in the given light wave
The magnetic field of the light wave is given by
$ B = B_0 \bigl[\sin(3.14 \times 10^7 \, c \, t) \;+\; \sin(6.28 \times 10^7 \, c \, t)\bigr]. $
There are two angular frequencies involved:
$ \omega_1 = 3.14 \times 10^7 \, c \quad \text{and} \quad \omega_2 = 6.28 \times 10^7 \, c. $
Since $\omega = 2\pi f$, the corresponding frequencies are
$ f_1 = \dfrac{\omega_1}{2\pi} \quad \text{and} \quad f_2 = \dfrac{\omega_2}{2\pi}. $
Step 2: Determine the maximum frequency
We want the frequency that produces the maximum kinetic energy in the photoelectrons.
Larger frequency $f_2$ corresponds to the second term with $\omega_2 = 6.28 \times 10^7 c.$
Thus,
$ f_{\text{max}} = \dfrac{6.28 \times 10^7 \, c}{2\pi}. $
Step 3: Calculate the maximum photon energy
The energy of a photon of frequency $f$ is given by $ E = h f. $ Hence,
$$
E_{\text{max}} = h \, f_{\text{max}}
= h \times \dfrac{\omega_2}{2\pi}
= h \times \dfrac{6.28 \times 10^7 \times c}{2\pi}.
$$
Substituting $h = 6.6 \times 10^{-34}\,\text{J·s}$ and $c = 3 \times 10^8\,\text{m/s}$:
$$
E_{\text{max}}
= 6.6 \times 10^{-34} \times \dfrac{6.28 \times 10^7 \times 3 \times 10^8}{2\pi} \,\text{J}.
$$
Step 4: Convert the photon energy to electronvolts
Recall that $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}.$
After simplifying, the expression typically gives a value around $12.4\,\text{eV}$ (a more precise calculation yields about $12.375\,\text{eV}$).
Thus,
$$
E_{\text{max}} \approx 12.375\,\text{eV}.
$$
Step 5: Subtract the work function to find maximum kinetic energy
The work function of silver is given as $ \phi = 4.7\,\text{eV}. $
The maximum kinetic energy of the photoelectrons is
$$
K_{\text{max}} = E_{\text{max}} - \phi
= 12.375\,\text{eV} - 4.7\,\text{eV}
\approx 7.675\,\text{eV}.
$$
Rounding properly, we get $ 7.72\,\text{eV} $ as the final answer.
