Determine The Wavelength Of Light Emitted When An Electron Falls From N 2 = 4 N_2 = 4 N 2 ​ = 4 To N 1 = 2 N_1 = 2 N 1 ​ = 2 (Balmer Series).2. Convert The Following Units Of Temperature: - 112°C To °F - 240°F To °C - 203°F To K - 360 K To °F3. Using

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Introduction

In this article, we will explore two distinct topics in physics and chemistry. First, we will determine the wavelength of light emitted when an electron falls from the fourth energy level (n2=4n_2 = 4) to the second energy level (n1=2n_1 = 2), which is part of the Balmer series. Then, we will convert various units of temperature from Celsius to Fahrenheit, Fahrenheit to Celsius, Fahrenheit to Kelvin, and Kelvin to Fahrenheit.

Determine the Wavelength of Light Emitted

Balmer Series

The Balmer series is a series of spectral lines in the visible region of the electromagnetic spectrum, which is emitted when an electron falls from a higher energy level to the second energy level (n1=2n_1 = 2). The energy of the electron is given by the formula:

En=−13.6 eVn2E_n = -\frac{13.6 \text{ eV}}{n^2}

where nn is the principal quantum number.

Energy Difference

To determine the wavelength of light emitted, we need to calculate the energy difference between the two energy levels. The energy of the electron in the fourth energy level (n2=4n_2 = 4) is:

E4=−13.6 eV42=−0.85 eVE_4 = -\frac{13.6 \text{ eV}}{4^2} = -0.85 \text{ eV}

The energy of the electron in the second energy level (n1=2n_1 = 2) is:

E2=−13.6 eV22=−3.4 eVE_2 = -\frac{13.6 \text{ eV}}{2^2} = -3.4 \text{ eV}

The energy difference between the two energy levels is:

ΔE=E4−E2=−0.85 eV−(−3.4 eV)=2.55 eV\Delta E = E_4 - E_2 = -0.85 \text{ eV} - (-3.4 \text{ eV}) = 2.55 \text{ eV}

Wavelength of Light Emitted

The wavelength of light emitted is related to the energy difference between the two energy levels by the formula:

λ=hcΔE\lambda = \frac{hc}{\Delta E}

where hh is Planck's constant, cc is the speed of light, and ΔE\Delta E is the energy difference between the two energy levels.

Plugging in the values, we get:

λ=(6.626×10−34 J s)(3×108 m/s)(2.55 eV)(1.602×10−19 J/eV)=4.56×10−7 m\lambda = \frac{(6.626 \times 10^{-34} \text{ J s}) (3 \times 10^8 \text{ m/s})}{(2.55 \text{ eV}) (1.602 \times 10^{-19} \text{ J/eV})} = 4.56 \times 10^{-7} \text{ m}

Converting this to nanometers, we get:

λ=456 nm\lambda = 456 \text{ nm}

Temperature Conversions

Convert 112°C to °F

To convert 112°C to °F, we use the formula:

T°F=T°C×95+32T_{\text{°F}} = T_{\text{°C}} \times \frac{9}{5} + 32

Plugging in the value, we get:

T°F=112×95+32=237.6 °FT_{\text{°F}} = 112 \times \frac{9}{5} + 32 = 237.6 \text{ °F}

Convert 240°F to °C

To convert 240°F to °C, we use the formula:

T°C=(T°F−32)×59T_{\text{°C}} = (T_{\text{°F}} - 32) \times \frac{5}{9}

Plugging in the value, we get:

T°C=(240−32)×59=115.56 °CT_{\text{°C}} = (240 - 32) \times \frac{5}{9} = 115.56 \text{ °C}

Convert 203°F to K

To convert 203°F to K, we use the formula:

TK=T°F×59+273.15T_{\text{K}} = T_{\text{°F}} \times \frac{5}{9} + 273.15

Plugging in the value, we get:

TK=203×59+273.15=286.11 KT_{\text{K}} = 203 \times \frac{5}{9} + 273.15 = 286.11 \text{ K}

Convert 360 K to °F

To convert 360 K to °F, we use the formula:

T°F=(TK−273.15)×95+32T_{\text{°F}} = (T_{\text{K}} - 273.15) \times \frac{9}{5} + 32

Plugging in the value, we get:

T°F=(360−273.15)×95+32=555.18 °FT_{\text{°F}} = (360 - 273.15) \times \frac{9}{5} + 32 = 555.18 \text{ °F}

Conclusion

Introduction

In our previous article, we explored two distinct topics in physics and chemistry: determining the wavelength of light emitted when an electron falls from the fourth energy level to the second energy level, and converting various units of temperature. In this article, we will answer some frequently asked questions related to these topics.

Q&A

Q: What is the Balmer series?

A: The Balmer series is a series of spectral lines in the visible region of the electromagnetic spectrum, which is emitted when an electron falls from a higher energy level to the second energy level (n1=2n_1 = 2).

Q: What is the formula for calculating the energy of an electron in a particular energy level?

A: The energy of an electron in a particular energy level is given by the formula:

En=−13.6 eVn2E_n = -\frac{13.6 \text{ eV}}{n^2}

where nn is the principal quantum number.

Q: How do I calculate the energy difference between two energy levels?

A: To calculate the energy difference between two energy levels, you need to subtract the energy of the lower energy level from the energy of the higher energy level.

Q: What is the formula for calculating the wavelength of light emitted?

A: The wavelength of light emitted is related to the energy difference between the two energy levels by the formula:

λ=hcΔE\lambda = \frac{hc}{\Delta E}

where hh is Planck's constant, cc is the speed of light, and ΔE\Delta E is the energy difference between the two energy levels.

Q: How do I convert Celsius to Fahrenheit?

A: To convert Celsius to Fahrenheit, you use the formula:

T°F=T°C×95+32T_{\text{°F}} = T_{\text{°C}} \times \frac{9}{5} + 32

Q: How do I convert Fahrenheit to Celsius?

A: To convert Fahrenheit to Celsius, you use the formula:

T°C=(T°F−32)×59T_{\text{°C}} = (T_{\text{°F}} - 32) \times \frac{5}{9}

Q: How do I convert Fahrenheit to Kelvin?

A: To convert Fahrenheit to Kelvin, you use the formula:

TK=T°F×59+273.15T_{\text{K}} = T_{\text{°F}} \times \frac{5}{9} + 273.15

Q: How do I convert Kelvin to Fahrenheit?

A: To convert Kelvin to Fahrenheit, you use the formula:

T°F=(TK−273.15)×95+32T_{\text{°F}} = (T_{\text{K}} - 273.15) \times \frac{9}{5} + 32

Q: What are some common applications of the Balmer series?

A: The Balmer series has several common applications, including:

  • Spectroscopy: The Balmer series is used to study the properties of atoms and molecules.
  • Astronomy: The Balmer series is used to study the properties of stars and other celestial objects.
  • Chemistry: The Balmer series is used to study the properties of chemical compounds.

Q: What are some common applications of temperature conversions?

A: Temperature conversions have several common applications, including:

  • Cooking: Temperature conversions are used to convert temperatures from Celsius to Fahrenheit and vice versa.
  • Science: Temperature conversions are used to convert temperatures from one unit to another.
  • Engineering: Temperature conversions are used to convert temperatures from one unit to another.

Conclusion

In this article, we answered some frequently asked questions related to determining the wavelength of light emitted when an electron falls from the fourth energy level to the second energy level, and converting various units of temperature. We hope that this article has been helpful in clarifying some of the concepts related to these topics.