Two energy levels of an electron in an atom are separated by 2.3 eV. The frequency of radiation emitted when the electrons go from higher to lower level is

Options

A. `6.95xx10^(14)Hz`
B. `3.68xx10^(15)Hz`
C. `5.6xx10^(14)Hz`
D. `9.11xx10^(15)Hz`

Answer

To find the frequency of radiation emitted when an electron transitions from a higher energy level to a lower energy level, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Difference**:
The energy difference between the two energy levels is given as \( \Delta E = 2.3 \, \text{eV} \).

2. **Convert Energy from eV to Joules**:
We need to convert the energy from electron volts (eV) to joules (J) because the Planck constant is in joules.
The conversion factor is:
\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \]
Therefore,
\[ \Delta E = 2.3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.68 \times 10^{-19} \, \text{J} \]

3. **Use Planck's Equation**:
The energy of the emitted radiation can also be expressed using Planck's equation:
\[ E = h \nu \]
where \( E \) is the energy, \( h \) is Planck's constant, and \( \nu \) is the frequency.

The value of Planck's constant is:
\[ h = 6.63 \times 10^{-34} \, \text{J s} \]

4. **Calculate the Frequency**:
Rearranging Planck's equation to find frequency gives:
\[ \nu = \frac{E}{h} \]
Substituting the values we have:
\[ \nu = \frac{3.68 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J s}} \]

5. **Perform the Calculation**:
\[ \nu \approx 5.55 \times 10^{14} \, \text{Hz} \]

### Final Answer:
The frequency of radiation emitted when the electron goes from a higher to a lower energy level is approximately \( 5.55 \times 10^{14} \, \text{Hz} \).