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 follow the details provided in the video transcript and apply the relevant physics concepts.

### Step 1: Understand the Work Functions
The work function of metal A, denoted as \( \phi_A \), is equal to the ionization energy of the hydrogen atom in its first excited state (n=2). The ionization energy for hydrogen in the first excited state can be calculated using the formula:

\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]

For hydrogen (\( Z = 1 \)) in the first excited state (\( n = 2 \)):

\[ \phi_A = -E_2 = -\left(-\frac{13.6 \cdot 1^2}{2^2}\right) = \frac{13.6}{4} = 3.4 \, \text{eV} \]

### Step 2: Calculate the Work Function for Metal B
The work function of metal B, denoted as \( \phi_B \), is equal to the ionization energy of the \( He^+ \) ion in the second orbit (n=2). Using the same formula:

For \( He^+ \) (\( Z = 2 \)) in the second orbit (\( n = 2 \)):

\[ \phi_B = -E_2 = -\left(-\frac{13.6 \cdot 2^2}{2^2}\right) = \frac{13.6 \cdot 2^2}{2^2} = 13.6 \, \text{eV} \]

### Step 3: Apply Einstein's Photoelectric Equation
According to Einstein's photoelectric equation, the maximum kinetic energy (\( K \)) of the emitted photoelectrons can be expressed as:

\[ K_A = E - \phi_A \]
\[ K_B = E - \phi_B \]

### Step 4: Relate the Kinetic Energies
From the problem statement, we know that the maximum kinetic energy of photoelectrons emitted from A is twice that of those emitted from B:

\[ K_A = 2K_B \]

Substituting the expressions for \( K_A \) and \( K_B \):

\[ E - \phi_A = 2(E - \phi_B) \]

### Step 5: Substitute the Values of Work Functions
Substituting \( \phi_A = 3.4 \, \text{eV} \) and \( \phi_B = 13.6 \, \text{eV} \):

\[ E - 3.4 = 2(E - 13.6) \]

### Step 6: Solve for E
Expanding the equation:

\[ E - 3.4 = 2E - 27.2 \]

Rearranging gives:

\[ E - 2E = -27.2 + 3.4 \]
\[ -E = -23.8 \]
\[ E = 23.8 \, \text{eV} \]

### Final Answer
The value of \( E \) is \( 23.8 \, \text{eV} \).

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