Question
What is the maximum wavelength of light required to remove an electron from a hydrogen atom in the ground state (n=1)?
Answers
GPT-4o mini
To find the maximum wavelength of light required to remove an electron from a hydrogen atom in the ground state, we need to calculate the ionization energy of the hydrogen atom when the electron is in the ground state (n=1).
The ionization energy \( E \) for a hydrogen atom can be calculated using the formula:
\[
E = -\frac{13.6 \, \text{eV}}{n^2}
\]
For the ground state (\( n = 1 \)):
\[
E = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
To ionize the atom, we need to supply energy equal to the absolute value of the ionization energy:
\[
E = 13.6 \, \text{eV}
\]
Next, to find the maximum wavelength \( \lambda \) of light that can provide this energy, we can use the relationship between energy and wavelength given by:
\[
E = \frac{hc}{\lambda}
\]
Where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \),
- \( c \) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \),
- \( \lambda \) is the wavelength in meters.
First, we need to convert the energy from electron volts to joules. The conversion factor is \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \).
\[
E = 13.6 \, \text{eV} \times (1.602 \times 10^{-19} \, \text{J/eV}) = 2.179 \times 10^{-18} \, \text{J}
\]
Now, we can rearrange the formula for wavelength \( \lambda \):
\[
\lambda = \frac{hc}{E}
\]
Substituting the values for \( h \), \( c \), and \( E \):
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{2.179 \times 10^{-18} \, \text{J}}
\]
Calculating the values:
\[
\lambda = \frac{1.9878 \times 10^{-25} \, \text{J} \cdot \text{m}}{2.179 \times 10^{-18} \, \text{J}} \approx 9.12 \times 10^{-8} \, \text{m}
\]
Now converting meters to nanometers (1 m = \( 10^9 \) nm):
\[
\lambda \approx 9.12 \times 10^{-8} \, \text{m} \times 10^9 \, \text{nm/m} \approx 91.2 \, \text{nm}
\]
Thus, the maximum wavelength of light required to remove an electron from a hydrogen atom in the ground state is approximately:
\[
\lambda \approx 91.2 \, \text{nm}
\]
The ionization energy \( E \) for a hydrogen atom can be calculated using the formula:
\[
E = -\frac{13.6 \, \text{eV}}{n^2}
\]
For the ground state (\( n = 1 \)):
\[
E = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV}
\]
To ionize the atom, we need to supply energy equal to the absolute value of the ionization energy:
\[
E = 13.6 \, \text{eV}
\]
Next, to find the maximum wavelength \( \lambda \) of light that can provide this energy, we can use the relationship between energy and wavelength given by:
\[
E = \frac{hc}{\lambda}
\]
Where:
- \( E \) is the energy in joules,
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \),
- \( c \) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \),
- \( \lambda \) is the wavelength in meters.
First, we need to convert the energy from electron volts to joules. The conversion factor is \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \).
\[
E = 13.6 \, \text{eV} \times (1.602 \times 10^{-19} \, \text{J/eV}) = 2.179 \times 10^{-18} \, \text{J}
\]
Now, we can rearrange the formula for wavelength \( \lambda \):
\[
\lambda = \frac{hc}{E}
\]
Substituting the values for \( h \), \( c \), and \( E \):
\[
\lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{2.179 \times 10^{-18} \, \text{J}}
\]
Calculating the values:
\[
\lambda = \frac{1.9878 \times 10^{-25} \, \text{J} \cdot \text{m}}{2.179 \times 10^{-18} \, \text{J}} \approx 9.12 \times 10^{-8} \, \text{m}
\]
Now converting meters to nanometers (1 m = \( 10^9 \) nm):
\[
\lambda \approx 9.12 \times 10^{-8} \, \text{m} \times 10^9 \, \text{nm/m} \approx 91.2 \, \text{nm}
\]
Thus, the maximum wavelength of light required to remove an electron from a hydrogen atom in the ground state is approximately:
\[
\lambda \approx 91.2 \, \text{nm}
\]