The series limit wavelength of the Lyman series for the hydrogen atom is given by

Options

A. `1/R`
B. `4/R`
C. `9/R`
D. `16/R`

Answer

To find the series limit wavelength of the Lyman series for the hydrogen atom, we can follow these steps:

### Step 1: Understand the Concept of Series Limit Wavelength
The series limit wavelength refers to the shortest wavelength in a spectral series, which occurs when an electron transitions from an infinite energy level (n = ∞) to the ground state (n = 1). This transition represents the maximum energy difference, thus yielding the shortest wavelength.

### Step 2: Write the Energy Levels
The energy of an electron in a hydrogen atom at a given energy level n is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the ground state (n = 1):
\[ E_1 = -13.6 \, \text{eV} \]
For the infinite state (n = ∞):
\[ E_{\infty} = 0 \, \text{eV} \]

### Step 3: Calculate the Energy Difference
The energy difference between the two states (from n = ∞ to n = 1) is:
\[ \Delta E = E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \, \text{eV} \]

### Step 4: Relate 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,
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)),
- \( \lambda \) is the wavelength.

### Step 5: Solve for Wavelength
Substituting the energy difference into the equation:
\[ 13.6 \, \text{eV} = \frac{hc}{\lambda} \]

To convert eV to Joules (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ 13.6 \, \text{eV} = 13.6 \times 1.6 \times 10^{-19} \, \text{J} = 2.176 \times 10^{-18} \, \text{J} \]

Now, substituting \( E \) into the equation:
\[ 2.176 \times 10^{-18} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{\lambda} \]

Rearranging to solve for \( \lambda \):
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2.176 \times 10^{-18}} \]

### Step 6: Calculate the Wavelength
Calculating the above expression gives:
\[ \lambda \approx 1.215 \times 10^{-7} \, \text{m} \]
or
\[ \lambda \approx 121.5 \, \text{nm} \]

### Final Answer
The series limit wavelength of the Lyman series for the hydrogen atom is approximately \( 121.5 \, \text{nm} \).

---