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.

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 \( n \) is given by the formula:

\[
E_n = -\frac{13.6 \text{ eV}}{n^2}
\]

where:
- \( E_n \) is the energy of the electron at level \( n \).
- \( 13.6 \text{ eV} \) is the ionization energy from the ground state (n = 1).

#### For \( n = 1 \):

\[
E_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6 \text{ eV}
\]

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

#### For \( n = 2 \):

\[
E_2 = -\frac{13.6 \text{ eV}}{2^2} = -\frac{13.6 \text{ eV}}{4} = -3.4 \text{ eV}
\]

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

### Step 2: Convert the energy to wavelength

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

\[
E = \frac{hc}{\lambda}
\]

where:
- \( E \) is the energy of the photon (in joules),
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ Js} \)),
- \( c \) is the speed of light in a vacuum (\( 3 \times 10^8 \text{ m/s} \)),
- \( \lambda \) is the wavelength of the photon (in meters).

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

\[
1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}
\]

So:

\[
\lambda = \frac{hc}{E}
\]

#### For \( n = 1 \):

\[
\lambda_1 = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{13.6 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV}}
\]

\[
\lambda_1 = \frac{1.988 \times 10^{-25} \text{ Js m}}{2.179 \times 10^{-18} \text{ J}} \approx 91.2 \text{ nm}
\]

#### For \( n = 2 \):

\[
\lambda_2 = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{3.4 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV}}
\]

\[
\lambda_2 = \frac{1.988 \times 10^{-25} \text{ Js m}}{5.447 \times 10^{-19} \text{ J}} \approx 364.7 \text{ nm}
\]

### Final Answers:
- **For \( n = 1 \)**: The maximum wavelength is approximately **91.2 nm**.
- **For \( n = 2 \)**: 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.

Discover how to calculate the maximum wavelength of light needed to ionize an electron from a hydrogen atom at energy levels n=1 and n=2. This physics problem explores key concepts like ionization energy, photon energy, and the energy-wavelength relationship, providing a step-by-step solution. Suitable for college-level students.

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