If an electron in a hydrogen atom jumps from the 3rd orbit to the 2 nd orbit, it emits `a` photon of wavelength `lambda`. When it jumps from the 4 th orbit to the 3 rd orbit, the corresponding wavelength of the photon will be `(m)/(n) lambda`. Find `(m+n)`.

Answer

To solve the problem, we will use the Rydberg formula for the wavelengths of emitted or absorbed photons when an electron transitions between energy levels in a hydrogen atom. The formula is given by:

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

where:
- \( \lambda \) is the wavelength of the emitted or absorbed photon,
- \( R_H \) is the Rydberg constant (\( R_H \approx 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) and \( n_2 \) are the principal quantum numbers of the lower and upper energy levels, respectively, with \( n_2 > n_1 \).

### Step 1: Calculate the wavelength for the transition from the 3rd orbit to the 2nd orbit

For the transition from the 3rd orbit (\( n_2 = 3 \)) to the 2nd orbit (\( n_1 = 2 \)):
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the terms:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) \]
Finding a common denominator (36):
\[ \frac{1}{\lambda} = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \]
Thus,
\[ \lambda = \frac{36}{5 R_H} \]

### Step 2: Calculate the wavelength for the transition from the 4th orbit to the 3rd orbit

For the transition from the 4th orbit (\( n_2 = 4 \)) to the 3rd orbit (\( n_1 = 3 \)):
\[ \frac{1}{\lambda'} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]
Calculating the terms:
\[ \frac{1}{\lambda'} = R_H \left( \frac{1}{9} - \frac{1}{16} \right) \]
Finding a common denominator (144):
\[ \frac{1}{\lambda'} = R_H \left( \frac{16 - 9}{144} \right) = R_H \left( \frac{7}{144} \right) \]
Thus,
\[ \lambda' = \frac{144}{7 R_H} \]

### Step 3: Relate the two wavelengths

We know that the wavelength for the transition from the 3rd to the 2nd orbit is \( \lambda = \frac{36}{5 R_H} \). Now we can express \( \lambda' \) in terms of \( \lambda \):
\[ \lambda' = \frac{144}{7 R_H} = \frac{144}{7} \cdot \frac{5}{36} \lambda \]
Calculating the fraction:
\[ \lambda' = \frac{144 \cdot 5}{7 \cdot 36} \lambda = \frac{720}{252} \lambda = \frac{120}{42} \lambda = \frac{20}{7} \lambda \]

### Step 4: Identify \( m \) and \( n \)

From the expression \( \lambda' = \frac{20}{7} \lambda \), we can identify \( m = 20 \) and \( n = 7 \).

### Step 5: Find \( m + n \)

Now, we calculate \( m + n \):
\[ m + n = 20 + 7 = 27 \]

Thus, the final answer is:
\[ \boxed{27} \]