A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits form the upper level to the lower level.

Options

A. `5.5xx10^(13)Hz`
B. `5.5xx10^(14)Hz`
C. `5.5xx10^(18)Hz`
D. `5.5xx10^(19)Hz`

Explanation

To solve the problem of finding the frequency of radiation emitted when an atom transitions from an upper energy level to a lower energy level, we will follow these steps:

### Step 1: Convert the energy difference from electron volts to joules
The energy difference between the two levels is given as 2.3 eV. We need to convert this energy into joules using the conversion factor:
1 eV = \(1.6 \times 10^{-19}\) joules.

\[ E = 2.3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \]
\[ E = 3.68 \times 10^{-19} \, \text{J} \]

### Step 2: Use the energy-frequency relationship
The relationship between energy (E) and frequency (ν) is given by Planck's equation:
\[ E = h \nu \]
where \(h\) is Planck's constant, approximately \(6.63 \times 10^{-34} \, \text{J s}\).

### Step 3: Rearrange the equation to solve for frequency
We can rearrange the equation to solve for frequency:
\[ \nu = \frac{E}{h} \]

### Step 4: Substitute the values into the equation
Now we can substitute the values of \(E\) and \(h\) into the equation:
\[ \nu = \frac{3.68 \times 10^{-19} \, \text{J}}{6.63 \times 10^{-34} \, \text{J s}} \]

### Step 5: Calculate the frequency
Now we perform the calculation:
\[ \nu \approx 5.55 \times 10^{14} \, \text{Hz} \]

### Conclusion
The frequency of radiation emitted when the atom transitions from the upper level to the lower level is approximately \(5.55 \times 10^{14} \, \text{Hz}\).

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