The work function of the metal A is equal to the ionization energy of the hydrogen atom in the first excited state. The work function of the metal B is equal to the ionization energy of `He^+` ion in the second orbit. Photons of the same energy E are incident on both A and B the maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B. Value of E (in eV) is

Options

A. 23.8 eV
B. 20.8 eV
C. 32.2 eV
D. 24.6 eV

Answer

To solve the problem step by step, we will first identify the work functions of metals A and B, then use the relationship between the maximum kinetic energies of the emitted photoelectrons to find the energy of the incident photons.

### Step 1: Determine the Work Function of Metal A
The work function of metal A is equal to the ionization energy of the hydrogen atom in the first excited state (n=2). The ionization energy for hydrogen in the nth orbit is given by:
\[ E_n = \frac{13.6 \, \text{eV}}{n^2} \]
For n=2:
\[ E_2 = \frac{13.6 \, \text{eV}}{2^2} = \frac{13.6 \, \text{eV}}{4} = 3.4 \, \text{eV} \]
Thus, the work function of metal A (\( \phi_A \)) is:
\[ \phi_A = 3.4 \, \text{eV} \]

### Step 2: Determine the Work Function of Metal B
The work function of metal B is equal to the ionization energy of the \( He^+ \) ion in the second orbit (n=2). For \( He^+ \) (where Z=2):
\[ E_n = \frac{Z^2 \cdot 13.6 \, \text{eV}}{n^2} \]
For n=2:
\[ E_2 = \frac{2^2 \cdot 13.6 \, \text{eV}}{2^2} = \frac{4 \cdot 13.6 \, \text{eV}}{4} = 13.6 \, \text{eV} \]
Thus, the work function of metal B (\( \phi_B \)) is:
\[ \phi_B = 13.6 \, \text{eV} \]

### Step 3: Relate the Maximum Kinetic Energies of Photoelectrons
The maximum kinetic energy of the photoelectrons emitted from metals A and B can be expressed as:
\[ K_A = E - \phi_A \]
\[ K_B = E - \phi_B \]
According to the problem, the maximum kinetic energy of photoelectrons emitted from A is twice that of those emitted from B:
\[ K_A = 2 K_B \]

### Step 4: Substitute the Expressions for Kinetic Energies
Substituting the expressions for \( K_A \) and \( K_B \):
\[ E - \phi_A = 2(E - \phi_B) \]

### Step 5: Substitute the Work Functions
Now substituting the values of \( \phi_A \) and \( \phi_B \):
\[ E - 3.4 = 2(E - 13.6) \]

### Step 6: Solve for E
Expanding the equation:
\[ E - 3.4 = 2E - 27.2 \]
Rearranging gives:
\[ E - 2E = -27.2 + 3.4 \]
\[ -E = -23.8 \]
Thus:
\[ E = 23.8 \, \text{eV} \]

### Final Answer
The value of \( E \) is:
\[ \boxed{23.8 \, \text{eV}} \]