Question

1. when x - rays with a frequency of 3.0×10¹⁶ hz are emitted from an x - ray machine.

a. what is the wavelength of the x - rays?
λ = c/ν
3.00×10⁸ m/s
3.0×10¹⁶
=
b. what is the energy emitted by the x - rays?

1. how much energy is released from a uv ray from the sun with a wavelength of 3.4×10⁻⁸ m?
2. what is the wavelength of light with a quantum of energy of 2.8×10⁻¹⁸ j?
3. the wavelength is 7.00 km. what is the energy associated with the wave?
4. 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?

Explanation:

Step1: Recall relevant formulas

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

Step2: Solve 6(a)

Given $
u = 3.0\times10^{16}\ Hz$, using $\lambda=\frac{c}{
u}$, 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 $E = h
u$, with $h = 6.63\times10^{-34}\ J\cdot s$ and $
u = 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 $E = h
u$, we find $
u=\frac{E}{h}=\frac{2.8\times10^{-18}\ J}{6.63\times10^{-34}\ J\cdot s}\approx4.22\times10^{15}\ Hz$
Then using $\lambda=\frac{c}{
u}$, $\lambda=\frac{3.00\times10^{8}\ m/s}{4.22\times10^{15}\ Hz}\approx7.11\times10^{-8}\ m$

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: $\approx3.06\times10^{-19}\ J$