The energy of a photon of visible light with a maximum wavelength in eV isA) 1B) 1.6C) 3.2D) 7

To find the energy of a photon of visible light with a maximum wavelength, we can use the equation that relates the energy of a photon to its wavelength:
E=hcλE = \frac{hc}{\lambda}
Where:
• EE is the energy of the photon,
• hh is Planck's constant (6.626×10−34 J\cdotps6.626 \times 10^{-34} \, \text{J·s}),
• cc is the speed of light (3×108 m/s3 \times 10^8 \, \text{m/s}),
• λ\lambda is the wavelength of the light.
Step 1: Find the wavelength corresponding to the maximum visible light
The visible spectrum of light has wavelengths ranging from about 400 nm (violet) to 700 nm (red). The maximum wavelength for visible light is typically considered to be around 700 nm (which is red light).
Step 2: Convert the wavelength into meters
λ=700 nm=700×10−9 m\lambda = 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m}
Step 3: Substitute into the equation
Now, we substitute the values into the energy equation:
E=(6.626×10−34 J\cdotps)×(3×108 m/s)700×10−9 mE = \frac{(6.626 \times 10^{-34} \, \text{J·s}) \times (3 \times 10^8 \, \text{m/s})}{700 \times 10^{-9} \, \text{m}} E=1.9878×10−25700×10−9 JE = \frac{1.9878 \times 10^{-25}}{700 \times 10^{-9}} \, \text{J} E=2.84×10−19 JE = 2.84 \times 10^{-19} \, \text{J}
Step 4: Convert energy from joules to electron volts (eV)
1 eV = 1.602×10−19 J1.602 \times 10^{-19} \, \text{J}, so:
E=2.84×10−191.602×10−19 eVE = \frac{2.84 \times 10^{-19}}{1.602 \times 10^{-19}} \, \text{eV} E≈1.77 eVE \approx 1.77 \, \text{eV}
Conclusion:
The energy of a photon of visible light with a maximum wavelength of about 700 nm is approximately 1.77 eV, which is closest to the value of 1.6 eV.
Therefore, the correct answer is: B) 1.6 eV.