Question

6.) a photon has an energy of 4.00×10⁻¹⁹ j. find... (a) the frequency (b) the wavelength (d) the region of the em spectrum/type of radiation 7.) a bright line spectrum contains a line with a wavelength of 518 nm. determine... (a) the wavelength in meters (b) the frequency (c) the energy (d) the color 8.) cobalt - 60 is an artificial radioisotope that is produced in a nuclear reactor for use as a gamma ray source in the treatment of certain types of cancer. if the wavelength of the gamma radiation from a cobalt - 60 source is 1.00×10⁻³ nm, calculate the energy of a photon of this radiation.

Explanation:

Step1: Recall energy - frequency formula

The energy of a photon is given by $E = h
u$, where $E$ is energy, $h = 6.63\times10^{- 34}\text{ J}\cdot\text{s}$ is Planck's constant and $
u$ is frequency. For the photon with $E = 4.00\times10^{-19}\text{ J}$ in problem 6(A), we can solve for $
u$: $
u=\frac{E}{h}$.
$
u=\frac{4.00\times 10^{-19}\text{ J}}{6.63\times 10^{-34}\text{ J}\cdot\text{s}}\approx6.03\times 10^{14}\text{ Hz}$

Step2: Find wavelength using frequency - wavelength relation

The speed of light $c = 3.00\times10^{8}\text{ m/s}$, and $c=\lambda
u$, so $\lambda=\frac{c}{
u}$. For the photon in 6(A), $\lambda=\frac{3.00\times 10^{8}\text{ m/s}}{6.03\times 10^{14}\text{ Hz}}\approx4.98\times 10^{-7}\text{ m} = 498\text{ nm}$

Step3: Determine EM - spectrum region

A wavelength of $498\text{ nm}$ is in the visible light region. Blue - green light has wavelengths around this value.

Step4: For problem 7(A)

Convert $518\text{ nm}$ to meters. Since $1\text{ nm}=10^{-9}\text{ m}$, then $\lambda = 518\times10^{-9}\text{ m}=5.18\times 10^{-7}\text{ m}$

Step5: For problem 7(B)

Using $c = \lambda
u$, we solve for $
u$. $
u=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\text{ m/s}}{5.18\times 10^{-7}\text{ m}}\approx5.79\times 10^{14}\text{ Hz}$

Step6: For problem 7(C)

Using $E = h
u$, with $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ and $
u = 5.79\times 10^{14}\text{ Hz}$, $E=(6.63\times 10^{-34}\text{ J}\cdot\text{s})\times(5.79\times 10^{14}\text{ Hz})\approx3.84\times 10^{-19}\text{ J}$

Step7: For problem 7(D)

A wavelength of $518\text{ nm}$ corresponds to green light in the visible spectrum.

Step8: For problem 8

First convert the wavelength of gamma - radiation from $1.00\times 10^{-3}\text{ nm}$ to meters. $\lambda=1.00\times 10^{-3}\times10^{-9}\text{ m}=1.00\times 10^{-12}\text{ m}$
Then use $c = \lambda
u$ to find $
u=\frac{c}{\lambda}=\frac{3.00\times 10^{8}\text{ m/s}}{1.00\times 10^{-12}\text{ m}} = 3.00\times 10^{20}\text{ Hz}$
Finally, use $E = h
u$ with $h = 6.63\times10^{-34}\text{ J}\cdot\text{s}$ to find $E=(6.63\times 10^{-34}\text{ J}\cdot\text{s})\times(3.00\times 10^{20}\text{ Hz})\approx1.99\times 10^{-13}\text{ J}$

Answer:

6(A). $6.03\times 10^{14}\text{ Hz}$
6(B). $498\text{ nm}$
6(D). Visible light (blue - green)
7(A). $5.18\times 10^{-7}\text{ m}$
7(B). $5.79\times 10^{14}\text{ Hz}$
7(C). $3.84\times 10^{-19}\text{ J}$
7(D). Green

1. $1.99\times 10^{-13}\text{ J}$