11. what is the energy of a 7.66×10¹⁴ hz wave?\n12. what is the frequency of a wave carrying 8.35×10⁻¹⁹ j of…

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by $E = h\nu$, where $E$ is energy, $h = 6.63\times10^{- 34}\text{ J}\cdot\text{s}$ (Planck's constant) and $\nu$ is frequency. First, we need to convert the wavelength $\lambda = 4.257\times10^{7}\text{ cm}$ to meters. Since $1\text{ m}=100\text{ cm}$, $\lambda=4.257\times10^{5}\text{ m}$. Then use the wave - speed formula $c = \lambda\nu$ ($c = 3\times10^{8}\text{ m/s}$), so $\nu=\frac{c}{\lambda}$. $\nu=\frac{3\times10^{8}\text{ m/s}}{4.257\times10^{5}\text{ m}}\approx704.7\text{ Hz}$ Then $E = h\nu=6.63\times10^{-34}\text{ J}\cdot\text{s}\times704.7\text{ Hz}\approx4.67\times10^{-31}\text{ J}$

Explanation:

Step1: Use energy - frequency formula

Given $\nu = 7.66\times10^{14}\text{ Hz}$, and $E = h\nu$ with $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$. $E=6.63\times10^{-34}\text{ J}\cdot\text{s}\times7.66\times10^{14}\text{ Hz}\approx5.08\times10^{-19}\text{ J}$

Explanation:

Step1: Rearrange energy - frequency formula

Given $E = 8.35\times10^{-19}\text{ J}$ and $E = h\nu$. We can solve for $\nu$ by $\nu=\frac{E}{h}$. $\nu=\frac{8.35\times10^{-19}\text{ J}}{6.63\times10^{-34}\text{ J}\cdot\text{s}}\approx1.26\times10^{15}\text{ Hz}$

Explanation:

Step1: Use energy - frequency formula

Given $\nu = 1.78\times10^{15}\text{ Hz}$ and $E = h\nu$ with $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$. $E=6.63\times10^{-34}\text{ J}\cdot\text{s}\times1.78\times10^{15}\text{ Hz}\approx1.18\times10^{-18}\text{ J}$

Explanation:

Step1: Identify frequency

Given $\nu = 3.12\times10^{14}\text{ s}^{-1}$ (which is the same as Hz), and $E = h\nu$ with $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$. $E=6.63\times10^{-34}\text{ J}\cdot\text{s}\times3.12\times10^{14}\text{ Hz}\approx2.07\times10^{-19}\text{ J}$

Explanation:

Step1: First find frequency

Given $E = 1.31\times10^{-22}\text{ J}$, and $E = h\nu$. So $\nu=\frac{E}{h}=\frac{1.31\times10^{-22}\text{ J}}{6.63\times10^{-34}\text{ J}\cdot\text{s}}\approx1.98\times10^{11}\text{ Hz}$ Then use $c = \lambda\nu$ to find $\lambda$. $\lambda=\frac{c}{\nu}=\frac{3\times10^{8}\text{ m/s}}{1.98\times10^{11}\text{ Hz}}\approx1.52\times10^{-3}\text{ m} = 1.52\text{ mm}$

Explanation:

Step1: Use energy - frequency formula

Given $E = 7.65\times10^{-17}\text{ J}$ and $E = h\nu$. First find $\nu=\frac{E}{h}=\frac{7.65\times10^{-17}\text{ J}}{6.63\times10^{-34}\text{ J}\cdot\text{s}}\approx1.15\times10^{17}\text{ Hz}$ If $\lambda = 9330\text{ cm}=93.3\text{ m}$, using $c = \lambda\nu$, $\nu=\frac{c}{\lambda}=\frac{3\times10^{8}\text{ m/s}}{93.3\text{ m}}\approx3.22\times10^{6}\text{ Hz}$ $E = h\nu=6.63\times10^{-34}\text{ J}\cdot\text{s}\times3.22\times10^{6}\text{ Hz}\approx2.14\times10^{-27}\text{ J}$

Explanation:

Step1: First find frequency

Given $E = 1.528\times10^{-13}\text{ J}$, and $E = h\nu$. So $\nu=\frac{E}{h}=\frac{1.528\times10^{-13}\text{ J}}{6.63\times10^{-34}\text{ J}\cdot\text{s}}\approx2.30\times10^{20}\text{ Hz}$ Then use $c = \lambda\nu$ to find $\lambda$. $\lambda=\frac{c}{\nu}=\frac{3\times10^{8}\text{ m/s}}{2.30\times10^{20}\text{ Hz}}\approx1.30\times10^{-12}\text{ m}$

Answer:

11. Energy $\approx4.67\times10^{-31}\text{ J}$
12. Energy $\approx5.08\times10^{-19}\text{ J}$
13. Frequency $\approx1.26\times10^{15}\text{ Hz}$
14. Energy $\approx1.18\times10^{-18}\text{ J}$
15. Energy $\approx2.07\times10^{-19}\text{ J}$
16. Frequency $\approx1.98\times10^{11}\text{ Hz}$, Wavelength $\approx1.52\text{ mm}$
17. Energy for first part $\approx1.15\times10^{17}\text{ Hz}$ - related energy, Energy for second part $\approx2.14\times10^{-27}\text{ J}$
18. Wavelength $\approx1.30\times10^{-12}\text{ m}$