Question

1. what is the frequency of electromagnetic radiation with a wavelength of 650 nm? (c= 3.0 x 10^{17} nm/s)
2. what is the energy of the radiation in the previous problem? (h= 6.626 x 10^{-34} j)

Explanation:

Question 5

Step1: Recall the formula for electromagnetic wave

The relationship between the speed of light \( c \), wavelength \( \lambda \), and frequency \(
u \) is \( c=\lambda
u \), so we can get \(
u = \frac{c}{\lambda} \).

Step2: Substitute the given values

Given \( c = 3.0\times10^{17}\space nm/s \) and \( \lambda=650\space nm \), substitute into the formula:
\(
u=\frac{3.0\times 10^{17}\space nm/s}{650\space nm} \)
Calculate the value: \(
u=\frac{3.0\times 10^{17}}{650}\space s^{-1}\approx4.62\times 10^{14}\space s^{-1} \)

Step1: Recall the formula for photon energy

The energy of a photon \( E \) is given by the formula \( E = h
u \), where \( h \) is Planck's constant and \(
u \) is the frequency. We already found \(
u\approx4.62\times 10^{14}\space s^{-1} \) from question 5, and \( h = 6.626\times 10^{-34}\space J\cdot s \)

Step2: Substitute the values into the formula

\( E=h
u=(6.626\times 10^{-34}\space J\cdot s)\times(4.62\times 10^{14}\space s^{-1}) \)
Calculate the value: \( E\approx6.626\times4.62\times 10^{-34 + 14}\space J\approx3.06\times 10^{-19}\space J \)

Answer:

The frequency of the electromagnetic radiation is approximately \( 4.62\times 10^{14}\space Hz \) (or \( 4.62\times 10^{14}\space s^{-1} \))

Question 6