1. Calculate the energy of one photon for the following waves: a. A radio station broadcasting on a wavelength of 7.5 meters. b. An ultraviolet wave from the sun, with a wavelength... 1. Calculate the energy of one photon for the following waves: a. A radio station broadcasting on a wavelength of 7.5 meters. b. An ultraviolet wave from the sun, with a wavelength of 6.8 nm. c. A wave from your laptop screen, with a wavelength of 472 nm. d. A wave from an alien's gamma ray gun, with a wavelength of 3 Angstroms (10^-10 M). 2. Find the wavelength (in nm) of a photon emitted from a Hydrogen electron transitioning from n = 4 to n = 2. 3. Find the de Broglie wavelength of the following objects: a. A 150 g cricket ball traveling at a speed of 40 m/s. b. Uranus, which has a weight of 8.6x10^25 kg and a speed of 24607 km/h. c. An electron moving at 6.7x10^6 m/s. d. Snickers the cat, who weighs 6 lbs and moves at 20 mph (when food is being poured out). Read more

Steps to Solve

1. Calculate Photon Energy for Different Wavelengths

To find the energy $E$ of a photon, we can use the formula:

$$ E = \frac{hc}{\lambda} $$

where:

• $h$ is Planck's constant ($6.626 \times 10^{-34} \text{ J s}$)
• $c$ is the speed of light ($3.00 \times 10^8 \text{ m/s}$)
• $\lambda$ is the wavelength (in meters)

a. For 7.5 m:

Convert the wavelength to meters (already in meters).

$$ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{7.5} $$

b. For 6.8 nm:

Convert 6.8 nm to meters:

$$ 6.8 \text{ nm} = 6.8 \times 10^{-9} \text{ m} $$

Then,

$$ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{6.8 \times 10^{-9}} $$

c. For 472 nm:

Convert 472 nm to meters:

$$ 472 \text{ nm} = 472 \times 10^{-9} \text{ m} $$

Then,

$$ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{472 \times 10^{-9}} $$

d. For 3 Angstroms:

Convert 3 Angstroms to meters:

$$ 3 \text{ Angstroms} = 3 \times 10^{-10} \text{ m} $$

Then,

$$ E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{3 \times 10^{-10}} $$

2. Find the Wavelength of the Hydrogen Photon Transition

For an electron transitioning from $n=4$ to $n=2$, use the Rydberg formula:

$$ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $$

where $R_H \approx 1.097 \times 10^7 \text{ m}^{-1}$, $n_1 = 2$, and $n_2 = 4$.

Calculate:

$$ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) $$

3. Calculate De Broglie Wavelengths

The de Broglie wavelength $\lambda_{dB}$ is given by:

$$ \lambda_{dB} = \frac{h}{mv} $$

where:

• $m$ is the mass (in kg)
• $v$ is the velocity (in m/s)

a. For Cricket Ball:

Convert mass to kg:

$$ 150 \text{ g} = 0.15 \text{ kg} $$

Calculate:

$$ \lambda_{dB} = \frac{(6.626 \times 10^{-34})}{(0.15)(40)} $$

b. For Uranus:

Mass is $8.6 \times 10^{25}$ kg, speed is $24607 \text{ km/h}$. Convert speed:

$$ 24607 \text{ km/h} = \frac{24607 \times 1000}{3600} \text{ m/s} $$

Calculate:

$$ \lambda_{dB} = \frac{(6.626 \times 10^{-34})}{(8.6 \times 10^{25})(v)} $$

c. For Electron:

With $v = 6.7 \times 10^6 \text{ m/s}$ and mass $m \approx 9.11 \times 10^{-31} \text{ kg}$:

$$ \lambda_{dB} = \frac{(6.626 \times 10^{-34})}{(9.11 \times 10^{-31})(6.7 \times 10^6)} $$

d. For Snickers the Cat:

Convert weight to mass (6 lbs = 2.72 kg) and speed (20 mph to m/s):

$$ 20 \text{ mph} = \frac{20 \times 1609.34}{