name: \nperform the following calculations—must show work for credit:\n1. a photon of light has a frequency…

Explanation:

Step1: Recall Planck - Einstein relation for photon energy

The energy of a photon is given by $E = h\nu$, where $h = 6.626\times10^{-34}\ J\cdot s$ (Planck's constant) and $\nu$ is the frequency of the photon.

Step2: Calculate energy for the first photon

Given $\nu_1=2.68\times 10^{6}\ Hz$, then $E_1=h\nu_1=(6.626\times 10^{-34}\ J\cdot s)\times(2.68\times 10^{6}\ s^{-1})$. [ \begin{align*} E_1&=(6.626\times2.68)\times10^{-34 + 6}\ J\ &=17.75768\times10^{-28}\ J\ &\approx1.78\times 10^{-27}\ J \end{align*} ] This frequency corresponds to radio - wave radiation (since radio - wave frequencies are in the range of $3\ Hz$ to $300\ GHz$ and $2.68\times 10^{6}\ Hz$ is in this range).

Step3: Calculate frequency of blue light

We know that the speed of light $c = 3.00\times 10^{8}\ m/s$, and the relation $c=\lambda\nu$. Given $\lambda = 4.00\times 10^{-7}\ m$, then $\nu_2=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\ m/s}{4.00\times 10^{-7}\ m}=7.50\times 10^{14}\ Hz$.

Step4: Calculate energy of blue light

Using $E = h\nu$, with $h = 6.626\times 10^{-34}\ J\cdot s$ and $\nu_2 = 7.50\times 10^{14}\ Hz$, we have $E_2=h\nu_2=(6.626\times 10^{-34}\ J\cdot s)\times(7.50\times 10^{14}\ s^{-1})$. [ \begin{align*} E_2&=(6.626\times7.50)\times10^{-34 + 14}\ J\ &=49.695\times10^{-20}\ J\ &\approx4.97\times 10^{-19}\ J \end{align*} ] This frequency and energy correspond to visible light (blue light is part of the visible - light spectrum which has frequencies in the range of approximately $4.0\times 10^{14}\ Hz$ to $7.5\times 10^{14}\ Hz$).

Answer:

1. Energy of the first photon: $E_1\approx1.78\times 10^{-27}\ J$, type of radiation: radio - wave.
2. Frequency of blue light: $\nu_2 = 7.50\times 10^{14}\ Hz$, energy of blue light: $E_2\approx4.97\times 10^{-19}\ J$