Ground state energy of H-atom is `(-E_(1))`,t he velocity of photoelectrons emitted when photon of energy `E_(2)` strikes stationary `Li^(2+)` ion in ground state will be:

Options

A. (a) `v=sqrt((2(E_(p)-E))/(m))`
B. (b) `v=sqrt((2(E_(p)+9E))/(m))`
C. (c) `v=sqrt((2(E_(p)-9E))/(m))`
D. (d) `v=sqrt((2(E_(p)-3E))/(m))`

Answer

To solve the problem step by step, we need to find the velocity of photoelectrons emitted when a photon of energy \( E_2 \) strikes a stationary \( \text{Li}^{2+} \) ion in its ground state.

### Step 1: Determine the ground state energy of \( \text{Li}^{2+} \)

The ground state energy of a hydrogen-like atom can be calculated using the formula:

\[ E_n = -\frac{Z^2 \cdot E_1}{n^2} \]

where:
- \( Z \) is the atomic number (for lithium, \( Z = 3 \)),
- \( E_1 \) is the ground state energy of hydrogen (\( E_1 = 13.6 \, \text{eV} \)),
- \( n \) is the principal quantum number (for ground state, \( n = 1 \)).

Substituting the values:

\[ E_{\text{Li}^{2+}} = -\frac{3^2 \cdot 13.6}{1^2} = -\frac{9 \cdot 13.6}{1} = -122.4 \, \text{eV} \]

### Step 2: Calculate the kinetic energy of emitted photoelectrons

When a photon of energy \( E_2 \) strikes the \( \text{Li}^{2+} \) ion, the energy absorbed by the ion is used to overcome the ion's binding energy and the remaining energy is converted into kinetic energy of the emitted photoelectrons.

The kinetic energy (\( KE \)) of the emitted photoelectrons can be expressed as:

\[ KE = E_2 - E_{\text{Li}^{2+}} \]

Substituting the energy of \( \text{Li}^{2+} \):

\[ KE = E_2 - (-122.4) = E_2 + 122.4 \]

### Step 3: Relate kinetic energy to velocity

The kinetic energy of the emitted photoelectrons can also be expressed in terms of their velocity \( v \):

\[ KE = \frac{1}{2} mv^2 \]

where \( m \) is the mass of the electron. Setting the two expressions for kinetic energy equal gives:

\[ E_2 + 122.4 = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity \( v \)

Rearranging the equation to solve for \( v \):

\[ v^2 = \frac{2(E_2 + 122.4)}{m} \]

Taking the square root:

\[ v = \sqrt{\frac{2(E_2 + 122.4)}{m}} \]

### Final Expression

Thus, the velocity of the photoelectrons emitted when a photon of energy \( E_2 \) strikes the stationary \( \text{Li}^{2+} \) ion in the ground state is given by:

\[ v = \sqrt{\frac{2(E_2 + 122.4)}{m}} \]