6. when x - rays with a frequency of 3.0×10¹⁶ hz are emitted from an x - ray machine.\na. what is the…

6. when x - rays with a frequency of 3.0×10¹⁶ hz are emitted from an x - ray machine.\na. what is the wavelength of the x - rays?\nλ = c/ν\n3.00×10⁸ m/s\n3.0×10¹⁶\n=\nb. what is the energy emitted by the x - rays?\n7. how much energy is released from a uv ray from the sun with a wavelength of 3.4×10⁻⁸ m?\n8. what is the wavelength of light with a quantum of energy of 2.8×10⁻¹⁸ j?\n9. the wavelength is 7.00 km. what is the energy associated with the wave?\n10. when strontium salts, such as strontium nitride and strontium carbonate, are heated, they produce a red color around 650 nm. what is the quantum of energy that is emitted at this wavelength?

6. when x - rays with a frequency of 3.0×10¹⁶ hz are emitted from an x - ray machine.\na. what is the wavelength of the x - rays?\nλ = c/ν\n3.00×10⁸ m/s\n3.0×10¹⁶\n=\nb. what is the energy emitted by the x - rays?\n7. how much energy is released from a uv ray from the sun with a wavelength of 3.4×10⁻⁸ m?\n8. what is the wavelength of light with a quantum of energy of 2.8×10⁻¹⁸ j?\n9. the wavelength is 7.00 km. what is the energy associated with the wave?\n10. when strontium salts, such as strontium nitride and strontium carbonate, are heated, they produce a red color around 650 nm. what is the quantum of energy that is emitted at this wavelength?

Answer

Explanation:

Step1: Recall relevant formulas

The speed - of - light formula is $c = \lambda\nu$ (where $c = 3.00\times10^{8}\ m/s$ is the speed of light in a vacuum, $\lambda$ is the wavelength, and $\nu$ is the frequency), and the energy formula is $E=h\nu=\frac{hc}{\lambda}$ (where $h = 6.63\times10^{- 34}\ J\cdot s$ is Planck's constant).

Step2: Solve 6(a)

Given $\nu = 3.0\times10^{16}\ Hz$, using $\lambda=\frac{c}{\nu}$, we substitute the values: $\lambda=\frac{3.00\times10^{8}\ m/s}{3.0\times10^{16}\ Hz}=1.0\times10^{-8}\ m$

Step3: Solve 6(b)

Using $E = h\nu$, with $h = 6.63\times10^{-34}\ J\cdot s$ and $\nu = 3.0\times10^{16}\ Hz$ $E=(6.63\times10^{-34}\ J\cdot s)\times(3.0\times10^{16}\ Hz)=1.989\times10^{-17}\ J$

Step4: Solve 7

Given $\lambda = 3.4\times10^{-8}\ m$, using $E=\frac{hc}{\lambda}$ $E=\frac{(6.63\times10^{-34}\ J\cdot s)\times(3.00\times10^{8}\ m/s)}{3.4\times10^{-8}\ m}\approx5.85\times10^{-18}\ J$

Step5: Solve 8

Given $E = 2.8\times10^{-18}\ J$, first from $E = h\nu$, we find $\nu=\frac{E}{h}=\frac{2.8\times10^{-18}\ J}{6.63\times10^{-34}\ J\cdot s}\approx4.22\times10^{15}\ Hz$ Then using $\lambda=\frac{c}{\nu}$, $\lambda=\frac{3.00\times10^{8}\ m/s}{4.22\times10^{15}\ Hz}\approx7.11\times10^{-8}\ m$

Step6: Solve 9

Given $\lambda=7.00\ km = 7.00\times10^{3}\ m$, using $E=\frac{hc}{\lambda}$ $E=\frac{(6.63\times10^{-34}\ J\cdot s)\times(3.00\times10^{8}\ m/s)}{7.00\times10^{3}\ m}\approx2.84\times10^{-29}\ J$

Step7: Solve 10

Given $\lambda = 650\ nm=650\times10^{-9}\ m$, using $E=\frac{hc}{\lambda}$ $E=\frac{(6.63\times10^{-34}\ J\cdot s)\times(3.00\times10^{8}\ m/s)}{650\times10^{-9}\ m}\approx3.06\times10^{-19}\ J$

Answer:

6(a): $1.0\times10^{-8}\ m$ 6(b): $1.989\times10^{-17}\ J$ 7: $\approx5.85\times10^{-18}\ J$ 8: $\approx7.11\times10^{-8}\ m$ 9: $\approx2.84\times10^{-29}\ J$ 10: $\approx3.06\times10^{-19}\ J$