Math Problem Statement

Calculate the maximum wavelength of light capable of removing an electron for a hydrogen atom from the energy state characterized by n = 1, by n = 2.

Solution

To determine the maximum wavelength of light capable of removing an electron from a hydrogen atom from a given energy state, we need to calculate the energy required to ionize the electron from that state and then relate it to the wavelength using the relationship between energy and wavelength.

Step 1: Calculate the ionization energy for each state

The energy of an electron in a hydrogen atom at a particular energy level nn is given by the formula:

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

where:

  • EnE_n is the energy of the electron at level nn.
  • 13.6 eV13.6 \text{ eV} is the ionization energy from the ground state (n = 1).

For n=1n = 1:

E1=13.6 eV12=13.6 eVE_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6 \text{ eV}

The energy required to ionize the electron from n=1n = 1 (i.e., to remove it from the atom) is 13.6 eV13.6 \text{ eV}.

For n=2n = 2:

E2=13.6 eV22=13.6 eV4=3.4 eVE_2 = -\frac{13.6 \text{ eV}}{2^2} = -\frac{13.6 \text{ eV}}{4} = -3.4 \text{ eV}

The energy required to ionize the electron from n=2n = 2 is 3.4 eV3.4 \text{ eV}.

Step 2: Convert the energy to wavelength

The energy of a photon is related to its wavelength by the equation:

E=hcλE = \frac{hc}{\lambda}

where:

  • EE is the energy of the photon (in joules),
  • hh is Planck's constant (6.626×1034 Js6.626 \times 10^{-34} \text{ Js}),
  • cc is the speed of light in a vacuum (3×108 m/s3 \times 10^8 \text{ m/s}),
  • λ\lambda is the wavelength of the photon (in meters).

We need to convert the energy from electron volts (eV) to joules. The conversion factor is:

1 eV=1.602×1019 J1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}

So:

λ=hcE\lambda = \frac{hc}{E}

For n=1n = 1:

λ1=6.626×1034 Js×3×108 m/s13.6 eV×1.602×1019 J/eV\lambda_1 = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{13.6 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV}}

λ1=1.988×1025 Js m2.179×1018 J91.2 nm\lambda_1 = \frac{1.988 \times 10^{-25} \text{ Js m}}{2.179 \times 10^{-18} \text{ J}} \approx 91.2 \text{ nm}

For n=2n = 2:

λ2=6.626×1034 Js×3×108 m/s3.4 eV×1.602×1019 J/eV\lambda_2 = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{3.4 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV}}

λ2=1.988×1025 Js m5.447×1019 J364.7 nm\lambda_2 = \frac{1.988 \times 10^{-25} \text{ Js m}}{5.447 \times 10^{-19} \text{ J}} \approx 364.7 \text{ nm}

Final Answers:

  • For n=1n = 1: The maximum wavelength is approximately 91.2 nm.
  • For n=2n = 2: The maximum wavelength is approximately 364.7 nm.

These wavelengths correspond to the ultraviolet region of the electromagnetic spectrum.

Would you like more details on any part of this process, or do you have any further questions?

Related Questions:

  1. What is the significance of the Rydberg constant in these calculations?
  2. How does the ionization energy change for higher energy levels (n > 2)?
  3. What is the relationship between energy levels and the emission spectrum of hydrogen?
  4. How can we calculate the wavelength for transitions between two energy levels?
  5. What are the practical applications of knowing the ionization energies for different elements?

Tip:

Remember that shorter wavelengths correspond to higher energy photons, which are capable of ionizing atoms more effectively.

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Math Problem Analysis

Mathematical Concepts

Quantum Physics
Photon Energy
Ionization Energy
Wavelength-Energy Relationship

Formulas

E_n = -13.6 eV / n^2
E = hc / λ
1 eV = 1.602 x 10^-19 J

Theorems

Bohr's Model of the Atom
Energy-Wavelength Relationship

Suitable Grade Level

College/University