The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from `lamda_1 " to " lamda_2` The work function of the metal is

Options

A. `(hc)/(lamda_1lamda_2)(2lamda_2-lamda_1)`
B. `(hc)/(lamda_1lamda_2)(2lamda_1-lamda_2)`
C. `(hc)/(lamda_1lamda_2)(lamda_1+lamda_2)`
D. `(hc)/(lamda_1lamda_2)(lamda_1-lamda_2)`

Answer

To solve the problem, we need to analyze the relationship between the kinetic energy of photoelectrons, the work function of the metal, and the wavelengths of the incident radiation. Let's break it down step by step.

### Step 1: Understanding the Energy of Photons
The energy of a photon is given by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h \) is Planck's constant,
- \( c \) is the speed of light,
- \( \lambda \) is the wavelength of the incident radiation.

### Step 2: Setting Up the Equations
When a photon strikes the metal surface, its energy is used to overcome the work function \( W \) of the metal and to provide kinetic energy \( K \) to the emitted electron. Therefore, we can write:
\[ \frac{hc}{\lambda_1} = W + K_1 \]
for the first wavelength \( \lambda_1 \), and
\[ \frac{hc}{\lambda_2} = W + K_2 \]
for the second wavelength \( \lambda_2 \).

### Step 3: Relating Kinetic Energies
According to the problem, the kinetic energy of the most energetic photoelectrons is doubled when the wavelength changes from \( \lambda_1 \) to \( \lambda_2 \). Thus, we can express this relationship as:
\[ K_2 = 2K_1 \]

### Step 4: Substituting Kinetic Energy
Substituting \( K_2 \) in the second equation gives:
\[ \frac{hc}{\lambda_2} = W + 2K_1 \]

### Step 5: Rearranging the Equations
Now we have two equations:
1. \( \frac{hc}{\lambda_1} = W + K_1 \) (Equation 1)
2. \( \frac{hc}{\lambda_2} = W + 2K_1 \) (Equation 2)

From Equation 1, we can express \( K_1 \):
\[ K_1 = \frac{hc}{\lambda_1} - W \]

### Step 6: Substituting \( K_1 \) into Equation 2
Substituting \( K_1 \) into Equation 2:
\[ \frac{hc}{\lambda_2} = W + 2\left(\frac{hc}{\lambda_1} - W\right) \]
Expanding this gives:
\[ \frac{hc}{\lambda_2} = W + \frac{2hc}{\lambda_1} - 2W \]
\[ \frac{hc}{\lambda_2} = \frac{2hc}{\lambda_1} - W \]

### Step 7: Rearranging to Find Work Function
Rearranging the above equation to isolate \( W \):
\[ W = \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_2} \]
\[ W = hc\left(\frac{2}{\lambda_1} - \frac{1}{\lambda_2}\right) \]

### Step 8: Final Expression for Work Function
Factoring out \( hc \):
\[ W = hc\left(\frac{2\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right) \]

### Conclusion
Thus, the work function of the metal is given by:
\[ W = \frac{hc(2\lambda_2 - \lambda_1)}{\lambda_1 \lambda_2} \]