3. isotopes vs. atomic mass
a. define isotope in your own words.
b. explain why the atomic mass of the periodic table is not a whole number.
a. quantitative isotope problem
boron exists naturally as two isotopes, b - 10 and b - 11.
isotope mass (amu) natural abundance
b - 10 10.013 19.78%
b - 11 11.009 80.22%
5. calculate the average atomic mass of boron, showing all work and using correct significant figures.
part iii: the nature of light (wave - particle duality)
8. light relationships
a. write the equation that relates wavelength, frequency, and the speed of light, and define each symbol.
b. describe the relationship between wavelength and frequency in words.
9. photon energy comparison
the energy of a 500 nm photon is 4.0×10⁻¹⁹ j.
calculate the energy of a 250 nm photon and explain why the energy changes.
10. emission vs. absorption
which electronic transition in a hydrogen atom emits the most energy, and why?
- n = 4 → n = 2
- n = 7 → n = 3
- n = 2 → n = 4
- n = 3 → n = 7

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Accepted Answer

### Question 3
# Brief Explanations:
a. An isotope is an atom of the same element (same proton count) with a different number of neutrons, so it has a different mass number.
b. The atomic mass on the periodic table is a weighted average of the masses of all an element's naturally occurring isotopes, accounting for their relative abundances, so it is not a whole number.
# Answer:
a. Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons (different mass numbers).
b. Atomic mass is a weighted average of an element's naturally occurring isotopes' masses, so it is not a whole number.

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### Question 5
# Explanation:
## Step1: Convert % to decimals
$19.78\% = 0.1978$, $80.22\% = 0.8022$
## Step2: Calculate isotope contributions
$	ext{Contribution of B-10} = 0.1978 	imes 10.013 = 1.9805714$
$	ext{Contribution of B-11} = 0.8022 	imes 11.009 = 8.8314198$
## Step3: Sum contributions for average
$	ext{Average atomic mass} = 1.9805714 + 8.8314198$
# Answer:
$10.81$ amu

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### Question 8
# Brief Explanations:
a. The core equation relates the three properties of light, with each symbol representing a specific measurable quantity.
b. Wavelength and frequency have an inverse relationship dictated by the constant speed of light.
# Answer:
a. The equation is $c = \lambda
u$
- $c$ = speed of light in a vacuum ($3.00 	imes 10^8$ m/s, a constant)
- $\lambda$ = wavelength of the light (distance between wave crests, units like m or nm)
- $
u$ = frequency of the light (number of wave cycles per second, units of Hz or s⁻¹)
b. Wavelength and frequency are inversely proportional: as wavelength increases, frequency decreases, and vice versa, because the speed of light in a vacuum is a fixed constant.

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### Question 9
# Explanation:
## Step1: Relate energy to wavelength
Photon energy follows $E = \frac{hc}{\lambda}$, so $E \propto \frac{1}{\lambda}$.
## Step2: Compare wavelengths
$\frac{\lambda_1}{\lambda_2} = \frac{500\ 	ext{nm}}{250\ 	ext{nm}} = 2$, so $E_2 = 2E_1$.
## Step3: Calculate new energy
$E_2 = 2 	imes 4.0 	imes 10^{-19}\ 	ext{J}$
# Answer:
$8.0 	imes 10^{-19}$ J
Energy doubles because photon energy is inversely proportional to wavelength; halving the wavelength doubles the energy.

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### Question 10
# Brief Explanations:
Only emission transitions release energy (electron moves to a lower energy level). The energy difference between levels is largest for the transition with the biggest gap between initial and final $n$ values.
# Answer:
$n=7
ightarrow n=3$
This is the only emission transition (electron moves from a higher to lower energy level) with the largest energy gap between the initial and final principal quantum numbers, so it emits the most energy. The transitions $n=2
ightarrow n=4$ and $n=3
ightarrow n=7$ are absorption transitions (they absorb energy, not emit it), and $n=4
ightarrow n=2$ has a smaller energy gap than $n=7
ightarrow n=3$.

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QUESTION IMAGE

Question

Explanation:

Question 3

Step1: Convert % to decimals

Step2: Calculate isotope contributions

Step3: Sum contributions for average

Answer:

Question 5