In an inelastic collision an electron excites a hydrogen atom from its ground state to a M-Shell state. A second electron collides instantaneously with the excited hydrogen atom in the m-Shell state and ionizes it. At leas how much energy the second electron transfors to the atom is the M-shell state?

Options

A. `+3.4 eV`
B. `+1.51 eV`
C. ` -3.4 eV`
D. `-1.51 eV`

Explanation

To solve the problem, we need to determine the minimum energy that the second electron must transfer to ionize the hydrogen atom that has been excited to the M-shell state. Here are the steps to arrive at the solution:

### Step 1: Identify the Energy Levels
The hydrogen atom has different energy levels defined by the principal quantum number \( n \). The ground state corresponds to \( n = 1 \), the K-shell to \( n = 1 \), the L-shell to \( n = 2 \), and the M-shell to \( n = 3 \).

### Step 2: Calculate the Energy of the M-shell
Using the Bohr model, the energy of an electron in a hydrogen atom at a given energy level can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the M-shell, where \( n = 3 \):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} \approx -1.51 \, \text{eV} \]

### Step 3: Determine the Energy Required for Ionization
To ionize the hydrogen atom from the M-shell, we need to provide enough energy to overcome the negative energy of the M-shell state and bring the electron to zero energy (free state). The energy required for ionization from the M-shell is the absolute value of the energy at that level:
\[ \text{Energy required for ionization} = |E_3| = 1.51 \, \text{eV} \]

### Step 4: Conclusion
Thus, the minimum energy that the second electron must transfer to the atom in the M-shell state to ionize it is approximately \( 1.51 \, \text{eV} \).

### Final Answer
The minimum energy the second electron must transfer to the atom in the M-shell state is **1.51 eV**.

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