Answer :
Sure, let's go through each part of your question step-by-step:
1. Wavelength of Light in the Balmer Series (n₂=4 to n₁=2):
The Balmer series corresponds to transitions where the final energy level (n₁) is 2.
To find the wavelength of light emitted, you can use the Rydberg formula:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
where [tex]\( R \)[/tex] is the Rydberg constant, approximately [tex]\( 1.097373 \times 10^7 \, \text{m}^{-1} \)[/tex].
Plug in [tex]\( n_1 = 2 \)[/tex] and [tex]\( n_2 = 4 \)[/tex]:
[tex]\[
\frac{1}{\lambda} = 1.097373 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)
\][/tex]
Solve to find [tex]\( \lambda \)[/tex].
2. Temperature Conversion:
- 112°C to Fahrenheit (F):
[tex]\[
F = \left(\frac{9}{5} \times 112\right) + 32 = 233.6
\][/tex]
- 240°F to Celsius (C):
[tex]\[
C = \frac{5}{9} \times (240 - 32) = 115.6
\][/tex]
- 203°F to Kelvin (K):
[tex]\[
C = \frac{5}{9} \times (203 - 32) \approx 95
\][/tex]
[tex]\[
K = C + 273.15 \approx 368.15
\][/tex]
- 360K to Fahrenheit (F):
[tex]\[
C = 360 - 273.15 = 86.85
\][/tex]
[tex]\[
F = \left(\frac{9}{5} \times 86.85\right) + 32 \approx 188.33
\][/tex]
3. Nuclear Chemistry: Alpha, Beta, Positron Emission, and Electron Capture of Thorium:
- Alpha Emission: Thorium (Th) loses a helium nucleus:
[tex]\[
^{2384}_{92}\text{Th} \rightarrow ^{2380}_{90}\text{Ra} + ^{4}_{2}\text{He}
\][/tex]
- Beta Emission: Thorium undergoes a neutron to proton conversion, emitting an electron:
[tex]\[
^{2384}_{92}\text{Th} \rightarrow ^{2384}_{93}\text{Pa} + e^- + \bar{\nu}_e
\][/tex]
- Positron Emission: Thorium decays, converting a proton to a neutron:
[tex]\[
^{2384}_{92}\text{Th} \rightarrow ^{2384}_{91}\text{Ac} + e^+ + \nu_e
\][/tex]
- Electron Capture: Thorium captures an electron, converting a proton to a neutron:
[tex]\[
^{2384}_{92}\text{Th} + e^- \rightarrow ^{2384}_{91}\text{Ac} + \nu_e
\][/tex]
4. Definitions:
- Nuclear Chemistry: The branch of chemistry dealing with radioactive elements, nuclear processes, and nuclear properties.
- Radiation: The emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles.
- Spectrum: The characteristic way in which the components of light (or other radiation) are separated and arranged, such as visible light separated into colors by a prism.
- Atomic Line: Specific wavelengths of light emitted or absorbed by electrons in atoms as they transition between energy levels.
5. Energy of a Photon:
The energy [tex]\( E \)[/tex] of a photon can be calculated using the formula:
[tex]\[
E = h \times \text{frequency}
\][/tex]
where [tex]\( h \)[/tex] is Planck's constant ([tex]\(6.62607015 \times 10^{-34} \)[/tex] J·s).
Given frequency [tex]\(5.0 \times 10^{10} \, \text{s}^{-1}\)[/tex], the energy [tex]\(E\)[/tex] is [tex]\(3.313035075 \times 10^{-23} \, \text{J}\)[/tex].
6. Wavelength for Transition from n₂=3 to n₁=2 (Balmer Series):
Using the Rydberg formula with [tex]\( n_1 = 2 \)[/tex] and [tex]\( n_2 = 3 \)[/tex], the wavelength [tex]\( \lambda \)[/tex] is approximately [tex]\(656.112 \text{ nm}\)[/tex].
7. Initial Energy Level n₂ for a Wavelength of 97.3 nm in the Lyman Series:
For the Lyman series (transitions to [tex]\( n_1 = 1 \)[/tex]), given [tex]\( \lambda = 97.3 \text{ nm}\)[/tex]:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right)
\][/tex]
Rearranging to solve for [tex]\( n_2 \)[/tex], we find [tex]\( n_2 \)[/tex] is approximately [tex]\( 3.0 \)[/tex].
Hopefully, this breaks down the process and answers each part of the question clearly!
1. Wavelength of Light in the Balmer Series (n₂=4 to n₁=2):
The Balmer series corresponds to transitions where the final energy level (n₁) is 2.
To find the wavelength of light emitted, you can use the Rydberg formula:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]
where [tex]\( R \)[/tex] is the Rydberg constant, approximately [tex]\( 1.097373 \times 10^7 \, \text{m}^{-1} \)[/tex].
Plug in [tex]\( n_1 = 2 \)[/tex] and [tex]\( n_2 = 4 \)[/tex]:
[tex]\[
\frac{1}{\lambda} = 1.097373 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)
\][/tex]
Solve to find [tex]\( \lambda \)[/tex].
2. Temperature Conversion:
- 112°C to Fahrenheit (F):
[tex]\[
F = \left(\frac{9}{5} \times 112\right) + 32 = 233.6
\][/tex]
- 240°F to Celsius (C):
[tex]\[
C = \frac{5}{9} \times (240 - 32) = 115.6
\][/tex]
- 203°F to Kelvin (K):
[tex]\[
C = \frac{5}{9} \times (203 - 32) \approx 95
\][/tex]
[tex]\[
K = C + 273.15 \approx 368.15
\][/tex]
- 360K to Fahrenheit (F):
[tex]\[
C = 360 - 273.15 = 86.85
\][/tex]
[tex]\[
F = \left(\frac{9}{5} \times 86.85\right) + 32 \approx 188.33
\][/tex]
3. Nuclear Chemistry: Alpha, Beta, Positron Emission, and Electron Capture of Thorium:
- Alpha Emission: Thorium (Th) loses a helium nucleus:
[tex]\[
^{2384}_{92}\text{Th} \rightarrow ^{2380}_{90}\text{Ra} + ^{4}_{2}\text{He}
\][/tex]
- Beta Emission: Thorium undergoes a neutron to proton conversion, emitting an electron:
[tex]\[
^{2384}_{92}\text{Th} \rightarrow ^{2384}_{93}\text{Pa} + e^- + \bar{\nu}_e
\][/tex]
- Positron Emission: Thorium decays, converting a proton to a neutron:
[tex]\[
^{2384}_{92}\text{Th} \rightarrow ^{2384}_{91}\text{Ac} + e^+ + \nu_e
\][/tex]
- Electron Capture: Thorium captures an electron, converting a proton to a neutron:
[tex]\[
^{2384}_{92}\text{Th} + e^- \rightarrow ^{2384}_{91}\text{Ac} + \nu_e
\][/tex]
4. Definitions:
- Nuclear Chemistry: The branch of chemistry dealing with radioactive elements, nuclear processes, and nuclear properties.
- Radiation: The emission of energy as electromagnetic waves or as moving subatomic particles, especially high-energy particles.
- Spectrum: The characteristic way in which the components of light (or other radiation) are separated and arranged, such as visible light separated into colors by a prism.
- Atomic Line: Specific wavelengths of light emitted or absorbed by electrons in atoms as they transition between energy levels.
5. Energy of a Photon:
The energy [tex]\( E \)[/tex] of a photon can be calculated using the formula:
[tex]\[
E = h \times \text{frequency}
\][/tex]
where [tex]\( h \)[/tex] is Planck's constant ([tex]\(6.62607015 \times 10^{-34} \)[/tex] J·s).
Given frequency [tex]\(5.0 \times 10^{10} \, \text{s}^{-1}\)[/tex], the energy [tex]\(E\)[/tex] is [tex]\(3.313035075 \times 10^{-23} \, \text{J}\)[/tex].
6. Wavelength for Transition from n₂=3 to n₁=2 (Balmer Series):
Using the Rydberg formula with [tex]\( n_1 = 2 \)[/tex] and [tex]\( n_2 = 3 \)[/tex], the wavelength [tex]\( \lambda \)[/tex] is approximately [tex]\(656.112 \text{ nm}\)[/tex].
7. Initial Energy Level n₂ for a Wavelength of 97.3 nm in the Lyman Series:
For the Lyman series (transitions to [tex]\( n_1 = 1 \)[/tex]), given [tex]\( \lambda = 97.3 \text{ nm}\)[/tex]:
[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right)
\][/tex]
Rearranging to solve for [tex]\( n_2 \)[/tex], we find [tex]\( n_2 \)[/tex] is approximately [tex]\( 3.0 \)[/tex].
Hopefully, this breaks down the process and answers each part of the question clearly!
