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, we will follow these steps:

### Step 1: Understand the relationship between energy, wavelength, and work function
The energy of a photon can be expressed as:
\[ E = h \nu = \frac{hc}{\lambda} \]
where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the incident light.

### Step 2: Write the equation for the maximum kinetic energy of the photoelectron
The maximum kinetic energy (KE) of the photoelectron can be expressed as:
\[ KE = E - \phi \]
where \(\phi\) is the work function of the metal.

### Step 3: Set up equations for the two scenarios
1. For the first case, where the maximum kinetic energy is \(E\) and the wavelength is \(\lambda\):
\[ E = \frac{hc}{\lambda} - \phi \quad \text{(Equation 1)} \]

2. For the second case, where the maximum kinetic energy becomes \(4E\) and the wavelength is reduced to \(\frac{\lambda}{3}\):
\[ 4E = \frac{hc}{\frac{\lambda}{3}} - \phi = \frac{3hc}{\lambda} - \phi \quad \text{(Equation 2)} \]

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

### Step 5: Simplify the equation
Expanding the left side:
\[ \frac{4hc}{\lambda} - 4\phi = \frac{3hc}{\lambda} - \phi \]
Rearranging gives:
\[ \frac{4hc}{\lambda} - \frac{3hc}{\lambda} = 4\phi - \phi \]
This simplifies to:
\[ \frac{hc}{\lambda} = 3\phi \]

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

### Final Answer
The work function of the metal is:
\[ \phi = \frac{hc}{3\lambda} \]

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