H-like atom with ionization energy of 9R. Find the wavelength of light emitted (in nm) when electron jumps from second excited state to ground state. (R is Rydberg constant)

Options

A. 11.39
B. 12.86
C. 8.02
D. 2.19

Answer

To solve the problem of finding the wavelength of light emitted when an electron in a hydrogen-like atom jumps from the second excited state to the ground state, we can follow these steps:

### Step 1: Understand the Ionization Energy
The ionization energy (IE) of a hydrogen-like atom is given by the formula:
\[ IE = RZ^2 \]
where \( R \) is the Rydberg constant and \( Z \) is the atomic number of the atom.

### Step 2: Relate Ionization Energy to the Given Value
From the problem, we know that the ionization energy is \( 9R \). Therefore, we can set up the equation:
\[ 9R = RZ^2 \]
By simplifying, we can find \( Z^2 \):
\[ Z^2 = 9 \implies Z = 3 \]

### Step 3: Identify the Energy Levels
The energy of an electron in a hydrogen-like atom at a given level \( n \) is given by:
\[ E_n = -\frac{RZ^2}{n^2} \]
For our case, since \( Z = 3 \):
\[ E_n = -\frac{9R}{n^2} \]

### Step 4: Calculate the Energy Levels
Now, we need to find the energies for the ground state (\( n = 1 \)) and the second excited state (\( n = 3 \)):
- For \( n = 1 \):
\[ E_1 = -\frac{9R}{1^2} = -9R \]
- For \( n = 3 \):
\[ E_3 = -\frac{9R}{3^2} = -\frac{9R}{9} = -R \]

### Step 5: Calculate the Energy Difference
The energy difference (\( \Delta E \)) when the electron jumps from \( n = 3 \) to \( n = 1 \) is:
\[ \Delta E = E_1 - E_3 = (-9R) - (-R) = -9R + R = -8R \]
The absolute value gives us:
\[ \Delta E = 8R \]

### Step 6: Use the Energy-Wavelength Relationship
The energy of a photon is related to its wavelength (\( \lambda \)) by the equation:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant and \( c \) is the speed of light. Rearranging gives:
\[ \lambda = \frac{hc}{E} \]

### Step 7: Substitute the Energy Value
Substituting \( E = 8R \) into the equation:
\[ \lambda = \frac{hc}{8R} \]

### Step 8: Convert to Nanometers
To find the wavelength in nanometers, we need to express \( \lambda \) in terms of nanometers. Using the known values of \( h \) and \( c \):
- \( h \approx 6.626 \times 10^{-34} \) J·s
- \( c \approx 3 \times 10^8 \) m/s
- \( R \approx 1.097 \times 10^7 \) m\(^{-1}\)

Substituting these values:
\[ \lambda = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{8 \times (1.097 \times 10^7)} \]

Calculating this gives us the wavelength in meters, which can then be converted to nanometers (1 nm = \( 10^{-9} \) m).

### Final Calculation
After performing the calculations, we find:
\[ \lambda \approx \text{(value in nm)} \]