An electron collides with a dfixed hydrogen atom in its ground stat. Hydrogen atom gets excited and the collding electron loses all its kinetic energy. Consequently the hydrgogen atom may emits a photon corresponding to the alrgest wavelength of the Balmer series. The K.E. of colliding electron will be 24.2/N eV. Find the value of N.

Explanation

To solve the problem, we need to follow these steps:

### Step 1: Understand the Energy Transition
The electron collides with a hydrogen atom in its ground state (n=1). The hydrogen atom gets excited, and the electron loses all its kinetic energy. The photon emitted corresponds to the largest wavelength of the Balmer series, which occurs during the transition from n=3 to n=2.

### Step 2: Calculate the Energy of the Photon
The energy of the photon emitted during the transition from n=3 to n=2 can be calculated using the Rydberg formula:

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

For the transition from n=3 to n=2:
- \( n_1 = 2 \)
- \( n_2 = 3 \)

Substituting these values into the formula gives:

\[ \frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

Finding a common denominator (36):

\[ \frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right) \]

### Step 3: Relate Energy to Wavelength
The energy of the photon can also be expressed in terms of wavelength:

\[ E = \frac{hc}{\lambda} \]

Substituting for \(\lambda\):

\[ E = hc \cdot \frac{36}{5R} \]

### Step 4: Substitute Known Values
Using:
- \( h = 6.63 \times 10^{-34} \, \text{J s} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)
- \( R = 1.097 \times 10^7 \, \text{m}^{-1} \)

Substituting these values into the energy equation:

\[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)(36)}{5(1.097 \times 10^7)} \]

Calculating this gives:

\[ E \approx 3.03 \times 10^{-19} \, \text{J} \]

### Step 5: Convert Energy to Electron Volts
To convert joules to electron volts, we use the conversion factor \(1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\):

\[ E \approx \frac{3.03 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.89 \, \text{eV} \]

### Step 6: Relate Kinetic Energy to Photon Energy
The kinetic energy of the colliding electron is given as:

\[ K.E. = \frac{24.2}{N} \, \text{eV} \]

Setting this equal to the energy of the emitted photon:

\[ \frac{24.2}{N} = 1.89 \]

### Step 7: Solve for N
Rearranging gives:

\[ N = \frac{24.2}{1.89} \approx 12.78 \]

### Conclusion
Thus, the value of \(N\) is approximately **12.78**.

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