Question

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

Short answer

The maximum wavelength of light capable of removing an electron for a hydrogen atom from the energy state characterized by \(n=1\) is approximately 91.18 nm, and for the energy state characterized by \(n=2\) is approximately 364.74 nm.

Step-by-step solution

Find the energy of an electron in the hydrogen atom for energy states n=1 and n=2

To find the energy of an electron in a hydrogen atom, we will use the formula: \[E_n = -\frac{13.6 \, eV}{n^2}\] where \(E_n\) is the energy of the electron in the nth energy level, and n is the principal quantum number. For \(n=1\), calculate the energy: \[E_1 = -\frac{13.6 \, eV}{1^2} = -13.6 \, eV\] For \(n=2\), calculate the energy: \[E_2 = -\frac{13.6 \, eV}{2^2} = -\frac{13.6 \, eV}{4} = -3.4 \, eV\]

Calculate the energy required to remove the electron from each energy state

When an electron is completely removed from an atom, it goes to the energy state with an energy of 0 eV. To find the energy required to remove the electron from each energy state, we have to calculate the difference between the energies in that state and the energy at 0 eV. Energy required to remove the electron from the state \(n=1\): \[\Delta E_1 = E_{\infty} - E_1 = 0 \, eV - (-13.6 \, eV) = 13.6 \, eV\] Energy required to remove the electron from the state \(n=2\): \[\Delta E_2 = E_{\infty} - E_2 = 0 \, eV - (-3.4 \, eV) = 3.4 \, eV\]

Find the wavelength corresponding to the energy required for each energy state

The energy and wavelength of a photon are related through the equation: \[E = h\cdot c / \lambda\] where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(4.135667696 \times 10^{-15} eV\cdot s\)), \(c\) is the speed of light (\(3 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of the photon. Rearrange the equation to find the wavelength that corresponds to the energy: \[\lambda = \frac{h\cdot c}{E}\] For the energy state n=1: \[\lambda_1 = \frac{4.135667696 \times 10^{-15} eV\cdot s \cdot 3 \times 10^8 m/s}{13.6 \, eV} = 91.18 \times 10^{-9} m\] For the energy state n=2: \[\lambda_2 = \frac{4.135667696 \times 10^{-15} eV \cdot s \cdot 3 \times 10^8 m/s}{3.4 \, eV} = 364.74 \times 10^{-9} m\] So, the maximum wavelength of light capable of removing an electron for a hydrogen atom from the energy state characterized by \(n=1\) is approximately 91.18 nm, and for the energy state characterized by \(n=2\) is approximately 364.74 nm.

Key concepts

Hydrogen Atom Energy Levels

The hydrogen atom, being the simplest atom, consists of one proton and one electron. The electron exists in specific energy levels or "shells," represented by the principal quantum number, denoted as \( n \).

• Each energy level corresponds to a distinct energy state.
• The formula to find the energy of an electron in the hydrogen atom is \( E_n = -\frac{13.6 \, eV}{n^2} \), where \( n \) is the principal quantum number.
• For \( n=1 \) (ground state), the energy \( E_1 \) is \(-13.6 \, eV\).
• For \( n=2 \) (first excited state), the energy \( E_2 \) is \(-3.4 \, eV\).

Electrons require energy to move from a lower energy level to a higher one or to be completely removed from an atom, which corresponds to moving to an energy level of \( 0 \, eV \). This concept is crucial for understanding photoelectron emissions, where photons provide the required energy for these transitions.

Wavelength Calculation

To find the wavelength of light that can remove an electron from a hydrogen atom, we use the relationship between wavelength and energy.
The formula \( E = \frac{h\cdot c}{\lambda} \) connects energy \( E \), Planck's constant \( h \), the speed of light \( c \), and wavelength \( \lambda \).

• \( h \) is Planck's constant, approximately \( 4.135667696 \times 10^{-15} \, eV\cdot s \).
• \( c \) is the speed of light, \( 3 \times 10^8 \, m/s \).
• \( \lambda \) is the wavelength we want to determine.

By rearranging the equation to \( \lambda = \frac{h\cdot c}{E} \), we can calculate \( \lambda \) using the energy calculated for either the \( n=1 \) or \( n=2 \) energy state.
The results show that for \( n=1 \), the maximum wavelength is approximately 91.18 nm, and for \( n=2 \), it is approximately 364.74 nm. These wavelengths indicate the types of photons that can liberate electrons from these states.

Planck's Equation

Planck's equation plays a key role in the interactions between energy and electromagnetic waves, such as light.
This equation suggests quantized energy states, meaning energy is exchanged in discrete amounts called quanta.

• The equation itself is \( E = h \cdot f \), where \( E \) is the energy, \( h \) is Planck's constant, and \( f \) is the frequency of the electromagnetic radiation.
• Combining this with the speed of light equation \( c = \lambda \cdot f \) gives us \( E = \frac{h\cdot c}{\lambda} \).
• Planck's constant \( h \) is essential for calculating the energy carried by each photon.

This principle helps in understanding photon-driven processes like photoelectron emissions. The electrons absorb the energy of incoming photons, facilitating their transition between energy levels or their complete ejection from an atom.
This understanding aids in advancing technologies such as solar panels and quantum computing.