Suppose in an imginary world the angular momentum is quantized to be even integral multiples of `h//2 pi`. What is the longest possible wavelenght emitted by hydrogen atoms in visible range in such a world according to Bohr's model?

Explanation

To find the longest possible wavelength emitted by hydrogen atoms in the visible range in an imaginary world where angular momentum is quantized to be even integral multiples of \( \frac{h}{2\pi} \), we can follow these steps:

### Step 1: Understand the quantization of angular momentum
In this imaginary world, the angular momentum \( L \) of an electron in a hydrogen atom is given by:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is an even integer (2, 4, 6, ...).

### Step 2: Determine the energy levels
The energy of the electron in the nth level of hydrogen is given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Since \( n \) can only take even values, the possible values for \( n \) are 2, 4, 6, etc.

### Step 3: Calculate the energy difference for transitions
To find the longest wavelength emitted, we need to find the smallest energy difference between two energy levels. The smallest energy difference occurs between consecutive levels. Therefore, we will consider \( n_1 = 2 \) and \( n_2 = 4 \).

The energy difference \( \Delta E \) between these two levels is:
\[ \Delta E = E_{n_1} - E_{n_2} = \left(-\frac{13.6}{2^2}\right) - \left(-\frac{13.6}{4^2}\right) \]
Calculating this gives:
\[ \Delta E = -\frac{13.6}{4} + \frac{13.6}{16} = -3.4 + 0.85 = -2.55 \, \text{eV} \]

### Step 4: Relate energy to wavelength
Using the relation between energy and wavelength:
\[ E = \frac{hc}{\lambda} \]
we can rearrange this to find the wavelength:
\[ \lambda = \frac{hc}{E} \]
Substituting the values of \( h \) and \( c \):
- \( h = 4.135667696 \times 10^{-15} \, \text{eV s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

Using \( E = 2.55 \, \text{eV} \):
\[ \lambda = \frac{1242 \, \text{eV nm}}{2.55 \, \text{eV}} \approx 487.05 \, \text{nm} \]

### Step 5: Conclusion
The longest possible wavelength emitted by hydrogen atoms in the visible range in this imaginary world is approximately:
\[ \lambda \approx 487 \, \text{nm} \]