Questions: Calculate the energy and wavelength associated with a photon of light that has a frequency of 4.02 × 10^13 s^-1. E=hc / K E=ho h=6.626 × 10^-34 Js

Solution Steps

We are given the frequency of the photon: \[ \nu = 4.02 \times 10^{13} \, \text{s}^{-1} \]

We are also given Planck's constant: \[ h = 6.626 \times 10^{-34} \, \text{Js} \]

The energy \( E \) of a photon can be calculated using the formula: \[ E = h \nu \]

Substituting the given values: \[ E = (6.626 \times 10^{-34} \, \text{Js}) \times (4.02 \times 10^{13} \, \text{s}^{-1}) \]

\[ E = 2.663652 \times 10^{-20} \, \text{J} \]

Rounding to four significant digits: \[ E = 2.664 \times 10^{-20} \, \text{J} \]

The wavelength \( \lambda \) can be calculated using the speed of light \( c \) and the frequency \( \nu \): \[ \lambda = \frac{c}{\nu} \]

Where the speed of light \( c \) is: \[ c = 3.00 \times 10^8 \, \text{m/s} \]

Substituting the given values: \[ \lambda = \frac{3.00 \times 10^8 \, \text{m/s}}{4.02 \times 10^{13} \, \text{s}^{-1}} \]

\[ \lambda = 7.462686567 \times 10^{-6} \, \text{m} \]

Rounding to four significant digits: \[ \lambda = 7.463 \times 10^{-6} \, \text{m} \]

Final Answer