Constants: h=6.626times 10^-34Js, q=1.6times 10^-19C, c=3times 10^8m/s 1. A single photon in free space has wavelength of 0.4mu m 1.A. Its frequency in Hertz (Hz) is __ (30 puan) a. 1.2times 10^8Hz b 1.2times 10^14Hz C. 7.5times 10^14Hz d. 1.6times 10^8Hz 1.B. Its energy in electron volts (eV) is __ (30 puan) là. 4.96 eV b. 1.77 eV IC. 0.177 eV d. 3.1 eV . Based on the relationship between photon energy and wavelength, a useful guideline is that E(in electron volts)=K/n where lambda is expressed in nanometers . The value of K is __ (40 puan) I a. 2.208 b. 1243 IC. 1.243 d. 2208

1.A. The frequency of a photon can be calculated using the equation $c = \lambda f$, where $c$ is the speed of light, $\lambda$ is the wavelength, and $f$ is the frequency. Rearranging the equation, we have $f = \frac{c}{\lambda}$. Plugging in the given values, we have $f = \frac{3 \times 10^8 m/s}{0.4 \times 10^{-6} m} = 7.5 \times 10^{14} Hz$. Therefore, the correct answer is option c.<br /><br />1.B. The energy of a photon can be calculated using the equation $E = hf$, where $h$ is Planck's constant and $f$ is the frequency. Plugging in the given values, we have $E = 6.626 \times 10^{-34} Js \times 7.5 \times 10^{14} Hz = 4.9695 \times 10^{-19} J$. To convert this energy to electron volts, we divide by the charge of an electron ($1.6 \times 10^{-19} C$), resulting in $E = 3.1 eV$. Therefore, the correct answer is option d.<br /><br />The value of K in the equation $E = K/\lambda$ is the energy in electron volts when $\lambda$ is expressed in nanometers. This relationship is derived from the equation $E = hf$, where $f = c/\lambda$. Rearranging this equation, we have $E = hc/\lambda$. Since $h$ and $c$ are constants, the value of K is equal to $hc$, which is $6.626 \times 10^{-34} Js \times 3 \times 10^8 m/s = 1.9878 \times 10^{-25} J \cdot nm$. Therefore, the correct answer is option b.