A photon of wavelength `lamda` is absorbed by an electron confined to a box of length `sqrt(35hlamda//8mc)` . As a result, the electron makes a transition from state k = 1 to the state n Subsequently the electron transits from the state n to the state m by emitting a photon of wavelength `lamda = 1.75lamda` then

Options

A. `n = 4 :m = 2`
B. `n = 5 :m = 3`
C. `n = 6 :m = 4`
D. `n = 3 :m = 1`

Explanation

To solve the problem step by step, we need to analyze the transitions of the electron in the potential box and the corresponding energy changes due to the absorption and emission of photons.

### Step 1: Understand the system
We have an electron confined in a one-dimensional box of length \( L = \sqrt{\frac{35h\lambda}{8mc}} \). The electron transitions from state \( k = 1 \) to state \( n \) after absorbing a photon of wavelength \( \lambda \).

### Step 2: Calculate the energy of the absorbed photon
The energy of the absorbed photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength of the photon.

### Step 3: Determine the energy levels in the box
The energy levels of an electron in a one-dimensional box are given by:
\[ E_n = \frac{n^2 h^2}{8mL^2} \]
where \( n \) is the quantum number, \( m \) is the mass of the electron, and \( L \) is the length of the box.

### Step 4: Calculate the length of the box
Substituting the expression for \( L \):
\[ L = \sqrt{\frac{35h\lambda}{8mc}} \]
we can find \( L^2 \):
\[ L^2 = \frac{35h\lambda}{8mc} \]

### Step 5: Calculate the energy for states \( k = 1 \) and \( n \)
For \( k = 1 \):
\[ E_1 = \frac{1^2 h^2}{8mL^2} = \frac{h^2}{8m \cdot \frac{35h\lambda}{8mc}} = \frac{hc}{35\lambda} \]

For state \( n \):
\[ E_n = \frac{n^2 h^2}{8mL^2} = \frac{n^2 hc}{35\lambda} \]

### Step 6: Energy difference due to the transition
The energy difference when the electron transitions from \( k = 1 \) to \( n \) is:
\[ E_n - E_1 = \frac{n^2 hc}{35\lambda} - \frac{hc}{35\lambda} = \frac{(n^2 - 1)hc}{35\lambda} \]

### Step 7: Photon emission during transition from \( n \) to \( m \)
The electron then emits a photon of wavelength \( \lambda' = 1.75\lambda \). The energy of this emitted photon is:
\[ E' = \frac{hc}{\lambda'} = \frac{hc}{1.75\lambda} = \frac{4hc}{7\lambda} \]

### Step 8: Energy difference for transition from \( n \) to \( m \)
The energy difference for the transition from \( n \) to \( m \) is:
\[ E_n - E_m = \frac{m^2 hc}{35\lambda} - \frac{n^2 hc}{35\lambda} = \frac{(n^2 - m^2)hc}{35\lambda} \]

### Step 9: Set the energies equal
Since the emitted photon energy equals the energy difference:
\[ \frac{(n^2 - m^2)hc}{35\lambda} = \frac{4hc}{7\lambda} \]
Cancelling \( hc/\lambda \) from both sides:
\[ \frac{n^2 - m^2}{35} = \frac{4}{7} \]
Multiplying through by 35:
\[ n^2 - m^2 = 20 \]

### Step 10: Solve for \( n \) and \( m \)
We know \( n^2 - m^2 = (n - m)(n + m) = 20 \). We can find pairs of integers \( (n, m) \) that satisfy this equation.

### Step 11: Find \( n \)
From the previous steps, we know that the electron was excited to state \( n \). If we assume \( n = 6 \) (as calculated earlier), we can find \( m \):
\[ 6^2 - m^2 = 20 \implies 36 - m^2 = 20 \implies m^2 = 16 \implies m = 4 \]

### Conclusion
The electron transitions from state \( n = 6 \) to state \( m = 4 \).