Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is `lambda`. If energy becomes four times when wavelength is reduced to one thrid, then work function of the metal is

Options

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

Explanation

To solve the problem step by step, we will use the concepts of photoelectric effect and the equations associated with it.

### Step 1: Write down the equation for maximum kinetic energy of a photoelectron.
The maximum kinetic energy (K.E.) of a photoelectron can be expressed as:
\[ K.E. = \frac{hc}{\lambda} - \phi \]
where:
- \( K.E. \) is the maximum kinetic energy,
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident light,
- \( \phi \) is the work function of the metal.

### Step 2: Set up the first condition using the given information.
From the problem, we know that when the wavelength is \( \lambda \), the maximum kinetic energy is \( E \). Therefore, we can write:
\[ E = \frac{hc}{\lambda} - \phi \]
This is our **Equation 1**.

### Step 3: Set up the second condition with the new wavelength.
Now, when the wavelength is reduced to \( \frac{\lambda}{3} \), the maximum kinetic energy becomes four times the previous value, which is \( 4E \). We can express this as:
\[ 4E = \frac{hc}{\frac{\lambda}{3}} - \phi \]
This simplifies to:
\[ 4E = \frac{3hc}{\lambda} - \phi \]
This is our **Equation 2**.

### Step 4: Substitute \( E \) from Equation 1 into Equation 2.
From Equation 1, we can express \( E \) as:
\[ E = \frac{hc}{\lambda} - \phi \]
Now, substituting this into Equation 2 gives:
\[ 4\left(\frac{hc}{\lambda} - \phi\right) = \frac{3hc}{\lambda} - \phi \]

### Step 5: Expand and simplify the equation.
Expanding the left side:
\[ \frac{4hc}{\lambda} - 4\phi = \frac{3hc}{\lambda} - \phi \]

Now, rearranging the equation:
\[ \frac{4hc}{\lambda} - \frac{3hc}{\lambda} = 4\phi - \phi \]
This simplifies to:
\[ \frac{hc}{\lambda} = 3\phi \]

### Step 6: Solve for the work function \( \phi \).
Now, we can express the work function \( \phi \):
\[ \phi = \frac{hc}{3\lambda} \]

### Conclusion
Thus, the work function of the metal is:
\[ \phi = \frac{hc}{3\lambda} \]