Question

(a) calculate the energy of a single photon of light with a frequency of 3.00×10¹⁶ s⁻¹.
energy =

(b) calculate the energy of a single photon of violet light with a wavelength of 402 nm.
energy =

Explanation:

Step1: Recall the formula for photon energy

The formula for the energy of a photon is \(E = h
u\), where \(h = 6.626\times10^{-34}\space J\cdot s\) (Planck's constant) and \(
u\) is the frequency.
For part (a), given \(
u=3.00\times 10^{16}\space s^{-1}\), substitute into the formula:
\(E=(6.626\times 10^{-34}\space J\cdot s)\times(3.00\times 10^{16}\space s^{-1})\)

Step2: Calculate the energy for part (a)

Using the rule of exponents \(a^m\times a^n=a^{m + n}\), we have \(E = 6.626\times3.00\times10^{-34 + 16}\space J\)
\(E = 19.878\times10^{-18}\space J=1.99\times 10^{-17}\space J\)

Step3: For part (b), first find the frequency from the wavelength

The relationship between speed of light \(c\), wavelength \(\lambda\) and frequency \(
u\) is \(c=\lambda
u\), where \(c = 3.00\times10^{8}\space m/s\) and \(\lambda=402\space nm=402\times10^{-9}\space m\)
So, \(
u=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\space m/s}{402\times 10^{-9}\space m}\)
\(
u=\frac{3.00\times10^{8}}{402\times10^{-9}}\space s^{-1}\approx7.46\times 10^{14}\space s^{-1}\)

Step4: Calculate the energy for part (b)

Using \(E = h
u\) with \(h = 6.626\times10^{-34}\space J\cdot s\) and \(
u = 7.46\times 10^{14}\space s^{-1}\)
\(E=(6.626\times 10^{-34}\space J\cdot s)\times(7.46\times 10^{14}\space s^{-1})\)
\(E=6.626\times7.46\times10^{-34 + 14}\space J\)
\(E = 49.43\times10^{-20}\space J = 4.94\times10^{-19}\space J\)

Answer:

(a) \(1.99\times 10^{-17}\space J\)
(b) \(4.94\times10^{-19}\space J\)