The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest interger) is (a) 802 nm (b) 823 nm
(c ) 1882 nm (d) 1648 nm.

Explanation

To find the smallest wavelength in the infrared region of the hydrogen spectrum, we can use the Rydberg formula for hydrogen:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

where:
- \( R \) is the Rydberg constant,
- \( n_1 \) is the principal quantum number of the lower energy level,
- \( n_2 \) is the principal quantum number of the higher energy level.

### Step 1: Identify the values for the infrared region
In the infrared region, the smallest wavelength corresponds to the transition from \( n_2 = \infty \) to \( n_1 = 3 \). This is because the infrared series starts from \( n_1 = 3 \) and goes to higher levels.

### Step 2: Set up the Rydberg formula for the infrared transition
For the transition from \( n_2 = \infty \) to \( n_1 = 3 \):
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) \]
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{9} \right) \]

### Step 3: Calculate the wavelength
Now, we can express \( \lambda_{\text{min}} \):
\[ \lambda_{\text{min}} = \frac{9}{R} \]

### Step 4: Relate the maximum wavelength in the UV region to the Rydberg formula
From the problem, we know that the largest wavelength in the ultraviolet region is \( 122 \, \text{nm} \). This corresponds to the transition from \( n_1 = 2 \) to \( n_2 = 1 \):
\[ \frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = R \left( \frac{1}{4} - 1 \right) = R \left( -\frac{3}{4} \right) \]
Thus,
\[ \frac{1}{122} = R \left( -\frac{3}{4} \right) \]

### Step 5: Solve for R
From the above equation, we can solve for \( R \):
\[ R = -\frac{4}{3 \times 122} \]

### Step 6: Substitute R back into the equation for \( \lambda_{\text{min}} \)
Now, substituting \( R \) back into the equation for \( \lambda_{\text{min}} \):
\[ \lambda_{\text{min}} = 9 \times \left(-\frac{4}{3 \times 122}\right)^{-1} \]
Calculating this gives:
\[ \lambda_{\text{min}} = 9 \times \frac{3 \times 122}{4} = \frac{9 \times 366}{4} = \frac{3294}{4} = 823.5 \, \text{nm} \]

### Step 7: Round to the nearest integer
Rounding \( 823.5 \) to the nearest integer gives \( 824 \, \text{nm} \).

### Conclusion
The smallest wavelength in the infrared region of the hydrogen spectrum is approximately \( 823 \, \text{nm} \), which corresponds to option (b).