The wavelength of light emitted from second orbit to first orbits in a hydrogen atom is

Options

A. `1.215 xx 10^(-7)m`
B. `1.215 xx 10^(-5) m`
C. `1.215 xx 10^(-4) m`
D. `1.215 xx 10^(-3) m`

Explanation

To find the wavelength of light emitted when an electron transitions from the second orbit (n=2) to the first orbit (n=1) in a hydrogen atom, we can follow these steps:

### Step 1: Determine the Energy Levels
The energy levels of a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

- For \( n=1 \):
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

- For \( n=2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -3.4 \, \text{eV} \]

### Step 2: Calculate the Energy Difference
The energy emitted when the electron transitions from the second orbit to the first orbit is given by the difference in energy levels:
\[ \Delta E = E_1 - E_2 \]
Substituting the values:
\[ \Delta E = (-13.6 \, \text{eV}) - (-3.4 \, \text{eV}) = -13.6 + 3.4 = -10.2 \, \text{eV} \]
The negative sign indicates that energy is emitted.

### Step 3: Convert Energy to Joules
To convert the energy from electron volts to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ \Delta E = 10.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.632 \times 10^{-18} \, \text{J} \]

### Step 4: Use the Energy-Wavelength Relationship
The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.67 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)

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

### Step 5: Substitute Values
Substituting the values into the equation:
\[ \lambda = \frac{(6.67 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{1.632 \times 10^{-18} \, \text{J}} \]

### Step 6: Calculate the Wavelength
Calculating the above expression:
\[ \lambda = \frac{2.001 \times 10^{-25}}{1.632 \times 10^{-18}} \approx 1.225 \times 10^{-7} \, \text{m} \]
Converting to nanometers (1 m = \( 10^9 \) nm):
\[ \lambda \approx 122.5 \, \text{nm} \]

### Final Answer
The wavelength of light emitted from the second orbit to the first orbit in a hydrogen atom is approximately \( 1.225 \times 10^{-7} \, \text{m} \) or \( 122.5 \, \text{nm} \).

---