If radiation corresponding to second line of "Balmer series" of `Li^(+2)` ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be:

Options

A. (a) `2.55eV`
B. (b) `4.25eV`
C. (c) `11.25eV`
D. (d) `19.55eV`

Answer

To solve the problem step by step, we need to find the kinetic energy of the ejected electron after the radiation corresponding to the second line of the Balmer series of the \( Li^{2+} \) ion knocks it out from the first excited state of the hydrogen atom.

### Step 1: Identify the transition for the second line of the Balmer series
The second line of the Balmer series corresponds to a transition from \( n = 4 \) to \( n = 2 \).

### Step 2: Calculate the energy of the emitted radiation
The energy of the emitted radiation can be calculated using the formula:
\[ E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
For the \( Li^{2+} \) ion, \( Z = 3 \), \( n_1 = 2 \), and \( n_2 = 4 \):
\[ E = 13.6 \times 3^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]
Calculating this:
\[ E = 13.6 \times 9 \left( \frac{1}{4} - \frac{1}{16} \right) \]
\[ = 13.6 \times 9 \left( \frac{4 - 1}{16} \right) = 13.6 \times 9 \times \frac{3}{16} \]
\[ = \frac{13.6 \times 27}{16} \, \text{eV} \]

### Step 3: Calculate the ionization energy of the hydrogen atom
The energy required to remove an electron from the first excited state of the hydrogen atom (where \( n = 2 \)) is given by:
\[ E_{ionization} = 13.6 \, Z^2 \left( \frac{1}{n_1^2} \right) \]
For hydrogen, \( Z = 1 \) and \( n_1 = 2 \):
\[ E_{ionization} = 13.6 \times 1^2 \left( \frac{1}{2^2} \right) = 13.6 \times \frac{1}{4} = 3.4 \, \text{eV} \]

### Step 4: Calculate the kinetic energy of the ejected electron
The kinetic energy (KE) of the ejected electron can be found by subtracting the ionization energy from the energy of the emitted radiation:
\[ KE = E - E_{ionization} \]
Substituting the values we calculated:
\[ KE = \frac{13.6 \times 27}{16} - 3.4 \]
Calculating \( \frac{13.6 \times 27}{16} \):
\[ = \frac{367.2}{16} = 22.95 \, \text{eV} \]
Now, substituting back to find KE:
\[ KE = 22.95 - 3.4 = 19.55 \, \text{eV} \]

### Final Answer
The kinetic energy of the ejected electron is \( 19.55 \, \text{eV} \).

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