The maximum wavelength that a sample of hydrogen atoms can absorb is

Options

A. 912 A
B. 1216 A
C. 1028 A
D. Infinite

Answer

To find the maximum wavelength that a sample of hydrogen atoms can absorb, we can follow these steps:

### Step 1: Understand the relationship between energy and wavelength
The energy (E) of a photon is related to its wavelength (λ) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \( c \) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \( \lambda \) is the wavelength in meters.

### Step 2: Determine the transition with minimum energy
To find the maximum wavelength, we need to minimize the energy. The transition with the least energy for hydrogen occurs from the second energy level (n=2) to the first energy level (n=1).

### Step 3: Calculate the change in energy (ΔE)
The energy levels of hydrogen can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the transition from n=2 to n=1:
- Energy at n=2:
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]
- Energy at n=1:
\[ E_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]

Now, calculate the change in energy (ΔE):
\[ \Delta E = E_1 - E_2 = (-13.6) - (-3.4) = -13.6 + 3.4 = -10.2 \, \text{eV} \]

### Step 4: Use the energy-wavelength relationship to find λ
Using the relationship \( \Delta E = \frac{12400}{\lambda} \) (where λ is in angstroms), we can rearrange it to solve for λ:
\[ \lambda = \frac{12400}{\Delta E} \]
Substituting ΔE:
\[ \lambda = \frac{12400}{10.2} \]

### Step 5: Calculate λ
Now, compute the value:
\[ \lambda = \frac{12400}{10.2} \approx 1215.69 \, \text{Å} \]

### Conclusion
The maximum wavelength that a sample of hydrogen atoms can absorb is approximately **1215.69 Å**.

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