Find the frequency of the photon emitted when an electron in the hydrogen atom makes a transition from n=9 to n=4.

h= 6.626 *10^-34 J*s

1 eV= 1.602*10^-19 J

Group of answer choices

1.65 * 10^13 Hz

1.65 * 10^14 Hz

2.46 * 10^13 Hz

2.46 * 10^14 Hz

To find the frequency of the photon emitted when an electron in a hydrogen atom transitions from \( n=9 \) to \( n=4 \), we need to follow these steps:

1. **Calculate the energy difference (\(\Delta E\)) between the two levels**.
2. **Use the energy difference to find the frequency (\(f\))** of the photon using the relation \( E = h f \).

The energy levels of a hydrogen atom are given by the formula:
\[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \]

### Step 1: Calculate the energy levels for \( n=9 \) and \( n=4 \):

- For \( n=9 \):
  \[ E_9 = - \frac{13.6 \, \text{eV}}{9^2} = - \frac{13.6}{81} \, \text{eV} = -0.1679 \, \text{eV} \]

- For \( n=4 \):
  \[ E_4 = - \frac{13.6 \, \text{eV}}{4^2} = - \frac{13.6}{16} \, \text{eV} = -0.85 \, \text{eV} \]

### Step 2: Find the energy difference (\(\Delta E\)):

\[ \Delta E = E_4 - E_9 = -0.85 \, \text{eV} - (-0.1679 \, \text{eV}) = -0.85 + 0.1679 \, \text{eV} = -0.6821 \, \text{eV} \]

Since the energy emitted is negative, we take the absolute value:
\[ \Delta E = 0.6821 \, \text{eV} \]

### Step 3: Convert the energy difference from eV to Joules:

\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \]
\[ \Delta E = 0.6821 \, \text{eV} \times 1.602 \times 10^{-19} \, \text{J/eV} \]
\[ \Delta E = 1.0922 \times 10^{-19} \, \text{J} \]

### Step 4: Use \( E = h f \) to find the frequency \( f \):

\[ \Delta E = h f \]
\[ f = \frac{\Delta E}{h} \]
\[ f = \frac{1.0922 \times 10^{-19} \, \text{J}}{6.626 \times 10^{-34} \, \text{J} \cdot \text{s}} \]
\[ f = 1.648 \times 10^{14} \, \text{Hz} \]

Therefore, the frequency of the photon emitted is approximately \( 1.65 \times 10^{14} \, \text{Hz} \).

### Correct Answer:
1.65 \(\times\) 10\(^14\) Hz

Would you like further details or have any questions? Here are some related questions for further exploration:

1. How do the energy levels in the hydrogen atom change with the principal quantum number \( n \)?
2. What is the significance of the Rydberg constant in hydrogen spectral lines?
3. How would the frequency change if the transition was from \( n=5 \) to \( n=2 \)?
4. How does the Bohr model of the hydrogen atom explain spectral lines?
5. What are the units of the Planck constant and how is it used in quantum mechanics?
6. Can you calculate the wavelength of the emitted photon from the frequency?
7. How does the energy of photons relate to their frequency and wavelength?
8. What are the main differences between the energy levels of hydrogen and other elements?

**Tip:** When dealing with transitions in the hydrogen atom, the differences in energy levels can often be simplified using the Rydberg formula, which directly relates the principal quantum numbers of the initial and final states.

Learn how to calculate the frequency of the photon emitted when an electron in a hydrogen atom transitions from n=9 to n=4. Follow a detailed step-by-step solution using quantum mechanics principles and Planck's equation (E = h*f).

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