section 3 - light
show your work for 10, 11 & 12 to get credit
c = 3 x 10^8 m/s
h = 6.628 x 10^(-34) j·s
c = λv and e = hv
7. with light waves, what is relationship between frequency and wavelength?
8. what type of light has wavelengths that can be larger than a meter?
9. under what conditions would an atom emit a photon?
10. (2 points) uv light has a median frequency of 1 x 10^15 hz, what is its wavelength?
11. (2 points) a photon has 3.31×10^(-19) joules of energy, what is its wavelength and what is its color?
12. (2 points) how much energy does a photon of blue light have if its wavelength is 407 nm?

Question

Accepted Answer

### Question 7
# Explanation:
## Step1: Recall the wave - speed formula
The speed of light formula is $c = \lambda v$, where $c$ is the speed of light ($c = 3	imes10^{8}\ m/s$), $\lambda$ is the wavelength and $v$ is the frequency.
## Step2: Rearrange the formula
We can rewrite it as $v=\frac{c}{\lambda}$. Since $c$ is a constant, frequency $v$ and wavelength $\lambda$ are inversely proportional.
# Answer:
The frequency and wavelength of light waves are inversely proportional.

### Question 8
# Brief Explanations:
Radio waves have wavelengths that can be larger than a meter. They are part of the electromagnetic spectrum with long - wavelengths.
# Answer:
Radio waves

### Question 9
# Brief Explanations:
An atom emits a photon when an electron in a higher - energy orbital drops to a lower - energy orbital. The energy difference between the two orbitals is released as a photon.
# Answer:
When an electron transitions from a higher - energy orbital to a lower - energy orbital.

### Question 10
# Explanation:
## Step1: Use the formula $c = \lambda v$
We know $c = 3	imes10^{8}\ m/s$ and $v = 1	imes10^{15}\ Hz$. Rearranging for $\lambda$, we get $\lambda=\frac{c}{v}$.
## Step2: Substitute the values
$\lambda=\frac{3	imes10^{8}\ m/s}{1	imes10^{15}\ Hz}=3	imes 10^{-7}\ m$
# Answer:
$3	imes 10^{-7}\ m$

### Question 11
# Explanation:
## Step1: First, use $E = h v$ to find the frequency
We know $E = 3.31	imes10^{-19}\ J$ and $h=6.628	imes 10^{-34}\ J\cdot s$. From $E = h v$, we can find $v=\frac{E}{h}$.
$v=\frac{3.31	imes10^{-19}\ J}{6.628	imes 10^{-34}\ J\cdot s}=5	imes10^{14}\ Hz$
## Step2: Then, use $c = \lambda v$ to find the wavelength
$\lambda=\frac{c}{v}=\frac{3	imes10^{8}\ m/s}{5	imes10^{14}\ Hz}=6	imes10^{-7}\ m = 600\ nm$ (visible light). Looking at the visible - light spectrum, a wavelength of 600 nm corresponds to orange light.
# Answer:
Wavelength: $6	imes10^{-7}\ m$; Color: Orange

### Question 12
# Explanation:
## Step1: Convert the wavelength to meters
$\lambda = 407\ nm=407	imes10^{-9}\ m$
## Step2: Use $c = \lambda v$ to find the frequency
$v=\frac{c}{\lambda}=\frac{3	imes10^{8}\ m/s}{407	imes10^{-9}\ m}\approx7.37	imes10^{14}\ Hz$
## Step3: Use $E = h v$ to find the energy
$E = h v=6.628	imes 10^{-34}\ J\cdot s	imes7.37	imes10^{14}\ Hz\approx4.88	imes10^{-19}\ J$
# Answer:
$4.88	imes10^{-19}\ J$

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QUESTION IMAGE

Question

Explanation:

Question 7

Step1: Recall the wave - speed formula

Step2: Rearrange the formula

Answer:

Question 8