The KE of the photoelectrons is E when the incident wavelength is `(lamda)/(2)`. The KE becomes 2E when the incident wavelength is `(lamda)/(3)`. The work function of the metal is

Options

A. `(hc)/lamda`
B. `(2hc)/(lamda)`
C. `(3hc)/(lamda)`
D. `(hc)/(3lamda)`

Explanation

To find the work function of the metal, we can use the photoelectric effect equation, which relates the kinetic energy of the emitted photoelectrons to the energy of the incident photons and the work function of the material. The equation is given by:

\[ KE = E_{photon} - \phi \]

where:
- \( KE \) is the kinetic energy of the emitted photoelectrons,
- \( E_{photon} \) is the energy of the incident photon,
- \( \phi \) is the work function of the metal.

The energy of the incident photon can be expressed in terms of its wavelength \( \lambda \) as:

\[ E_{photon} = \frac{hc}{\lambda} \]

where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light.

### Step-by-step Solution:

1. **Identify the first scenario:**
- When the wavelength is \( \frac{\lambda}{2} \), the kinetic energy \( KE_1 = E \).
- The energy of the photon is:
\[ E_{photon1} = \frac{hc}{\frac{\lambda}{2}} = \frac{2hc}{\lambda} \]
- Using the photoelectric effect equation:
\[ E = \frac{2hc}{\lambda} - \phi \quad \text{(1)} \]

2. **Identify the second scenario:**
- When the wavelength is \( \frac{\lambda}{3} \), the kinetic energy \( KE_2 = 2E \).
- The energy of the photon is:
\[ E_{photon2} = \frac{hc}{\frac{\lambda}{3}} = \frac{3hc}{\lambda} \]
- Using the photoelectric effect equation:
\[ 2E = \frac{3hc}{\lambda} - \phi \quad \text{(2)} \]

3. **Set up the equations:**
- From equation (1):
\[ \phi = \frac{2hc}{\lambda} - E \]
- From equation (2):
\[ \phi = \frac{3hc}{\lambda} - 2E \]

4. **Equate the two expressions for the work function \( \phi \):**
\[ \frac{2hc}{\lambda} - E = \frac{3hc}{\lambda} - 2E \]

5. **Rearranging the equation:**
- Move all terms involving \( E \) to one side:
\[ 2E - E = \frac{3hc}{\lambda} - \frac{2hc}{\lambda} \]
- Simplifying gives:
\[ E = \frac{hc}{\lambda} \]

6. **Substituting \( E \) back into one of the equations for \( \phi \):**
- Using equation (1):
\[ \phi = \frac{2hc}{\lambda} - \frac{hc}{\lambda} \]
- Simplifying gives:
\[ \phi = \frac{hc}{\lambda} \]

### Final Answer:
The work function of the metal is:
\[ \phi = \frac{hc}{\lambda} \]