What is the wavelength of ligth for the least energetic photon emitted in the Lyman series of the hydrogen spectrum. (Take , hc =1240 eV -nm)

Options

A. 122nm
B. 82nm
C. 150nm
D. 120nm

Explanation

To find the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum, we can follow these steps:

### Step 1: Identify the Transition
The least energetic photon in the Lyman series corresponds to the transition from the second energy level (n=2) to the first energy level (n=1).

### Step 2: Calculate the Energies of the Levels
The energy of an electron in a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
- For \( n = 1 \):
\[ E_1 = -\frac{13.6}{1^2} = -13.6 \, \text{eV} \]
- For \( n = 2 \):
\[ E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

### Step 3: Calculate the Energy of the Photon
The energy of the photon emitted during the transition from n=2 to n=1 is given by:
\[ E = E_2 - E_1 \]
Substituting the values:
\[ E = (-3.4) - (-13.6) = -3.4 + 13.6 = 10.2 \, \text{eV} \]

### Step 4: Use the Energy-Wavelength Relation
The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength.

Rearranging this formula to solve for \( \lambda \):
\[ \lambda = \frac{hc}{E} \]

### Step 5: Substitute the Known Values
Given that \( hc = 1240 \, \text{eV} \cdot \text{nm} \) and \( E = 10.2 \, \text{eV} \):
\[ \lambda = \frac{1240 \, \text{eV} \cdot \text{nm}}{10.2 \, \text{eV}} \]

### Step 6: Calculate the Wavelength
Now, performing the calculation:
\[ \lambda = \frac{1240}{10.2} \approx 121.56 \, \text{nm} \]

### Step 7: Round the Result
Rounding the result gives:
\[ \lambda \approx 122 \, \text{nm} \]

### Final Answer
The wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum is approximately **122 nm**.

---