An electron jumps from an outer orbit to an inner orbit with the energy difference of 3.0 eV. What will be the wavelength of the line and in what region does the emission take place?

Options

A. 3660 Å
B. 3620 Å
C. 4140 Å
D. 4560 Å

Answer

To solve the problem of an electron jumping from an outer orbit to an inner orbit with an energy difference of 3.0 eV, we can follow these steps:

### Step 1: Convert Energy from eV to Joules
The energy difference (ΔE) is given as 3.0 eV. To convert this energy into joules, we use the conversion factor:
1 eV = 1.6 x 10^-19 Joules.

\[ \Delta E = 3.0 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.8 \times 10^{-19} \, \text{J} \]

### Step 2: Use the Energy-Wavelength Relationship
The relationship between energy and wavelength is given by the equation:
\[ \Delta E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (6.67 x 10^-34 J·s),
- \( c \) is the speed of light (3.0 x 10^8 m/s),
- \( \lambda \) is the wavelength in meters.

### Step 3: Rearranging the Equation to Solve for Wavelength
Rearranging the equation to solve for wavelength (\( \lambda \)):
\[ \lambda = \frac{hc}{\Delta E} \]

### Step 4: Substitute the Values
Now, substitute the values of \( h \), \( c \), and \( \Delta E \) into the equation:
\[ \lambda = \frac{(6.67 \times 10^{-34} \, \text{J·s}) \times (3.0 \times 10^8 \, \text{m/s})}{4.8 \times 10^{-19} \, \text{J}} \]

### Step 5: Calculate the Wavelength
Calculating the above expression:
\[ \lambda = \frac{2.001 \times 10^{-25}}{4.8 \times 10^{-19}} \approx 4.17 \times 10^{-7} \, \text{m} \]

### Step 6: Convert Wavelength to Angstroms
To convert meters to angstroms (1 angstrom = 10^-10 m):
\[ \lambda \approx 4.17 \times 10^{-7} \, \text{m} \times 10^{10} \, \text{Å/m} = 4170 \, \text{Å} \]

### Step 7: Determine the Emission Region
The wavelength of approximately 4170 Å falls within the ultraviolet (UV) region of the electromagnetic spectrum.

### Final Answer
The wavelength of the emitted photon is approximately 4170 Å, and the emission takes place in the ultraviolet region.

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