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**Essay:**

1. Determine the wavelength of light emitted when an electron falls from [tex]n_2 = 4[/tex] to [tex]n_1 = 2[/tex] (Balmer series).

2. Convert the following units of temperature:
- [tex]112^\circ C[/tex] to [tex]^\circ F[/tex]
- [tex]240^\circ F[/tex] to [tex]^\circ C[/tex]
- [tex]203^\circ F[/tex] to K
- [tex]360 \, \text{K}[/tex] to [tex]^\circ F[/tex]

3. With the help of nuclear chemistry, write a complete alpha, beta, and positron emission equation.

Answer :

Sure, let's break down the solution to each part of the question step by step:

1. Determine the Wavelength of Light Emitted (Balmer Series):

When an electron falls from a higher energy level to a lower one in the hydrogen atom, it emits light. The wavelength of this light for transitions in the Balmer series can be found using the Rydberg formula:

[tex]\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\][/tex]

- Here, [tex]\( R = 1.097 \times 10^7 \, \text{m}^{-1} \)[/tex] (Rydberg constant).
- [tex]\( n_2 = 4 \)[/tex] and [tex]\( n_1 = 2 \)[/tex] for the transition of interest.

Substitute the values to find [tex]\( \frac{1}{\lambda} \)[/tex]:

[tex]\[
\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)
\][/tex]

Calculate this to find [tex]\( \lambda \)[/tex]:

[tex]\[
\lambda = 4.861744150714068 \times 10^{-7} \, \text{m} = 486.174 \, \text{nm}
\][/tex]

So, the wavelength of the light is approximately 486.174 nm.

2. Convert the Following Units of Temperature:

- Convert 112°C to Fahrenheit:

Use the conversion formula:

[tex]\[
F = \left(\frac{9}{5} \times C\right) + 32
\][/tex]

Substituting [tex]\( C = 112 \)[/tex]:

[tex]\[
F = \left(\frac{9}{5} \times 112\right) + 32 = 233.6°F
\][/tex]

- Convert 240°F to Celsius:

Use the formula:

[tex]\[
C = \left(F - 32\right) \times \frac{5}{9}
\][/tex]

Substituting [tex]\( F = 240 \)[/tex]:

[tex]\[
C = (240 - 32) \times \frac{5}{9} = 115.56°C
\][/tex]

- Convert 203°F to Kelvin:

First, convert 203°F to Celsius:

[tex]\[
C = (203 - 32) \times \frac{5}{9} = 95°C
\][/tex]

Now, convert Celsius to Kelvin:

[tex]\[
K = C + 273.15
\][/tex]

[tex]\[
K = 95 + 273.15 = 368.15 \, K
\][/tex]

- Convert 360 K to Fahrenheit:

First, convert Kelvin to Celsius:

[tex]\[
C = K - 273.15
\][/tex]

[tex]\[
C = 360 - 273.15 = 86.85°C
\][/tex]

Now, convert Celsius to Fahrenheit:

[tex]\[
F = \left(\frac{9}{5} \times 86.85\right) + 32 = 188.33°F
\][/tex]

These detailed steps show how to determine the wavelength in the Balmer series and convert temperatures between different scales accurately.

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