A surface irradiated with light `lamda=480 nm` gives out electrons with maximum velocity v `ms^(-1)`, the cut off wavelength being 600 nm. The same surface would release electrons with maximum velocity `2vms^(-1)` if it is irradiated by light of wavelength.

Options

A. 325 nm
B. 360 nm
C. 384 nm
D. 300 nm

Answer

To solve the problem, we will use the principles of the photoelectric effect and Einstein's photoelectric equation. The steps are as follows:

### Step 1: Understand the given data
- Wavelength of light (\( \lambda_1 \)) = 480 nm
- Maximum velocity of electrons (\( v \)) when irradiated with \( \lambda_1 \)
- Cut-off wavelength (\( \lambda_2 \)) = 600 nm
- Maximum velocity of electrons when irradiated with a new wavelength = \( 2v \)

### Step 2: Apply Einstein's photoelectric equation
The photoelectric equation is given by:
\[ E = \phi + KE \]
Where:
- \( E \) is the energy of the incident photons, \( E = \frac{hc}{\lambda} \)
- \( \phi \) is the work function of the material
- \( KE \) is the kinetic energy of the emitted electrons, \( KE = \frac{1}{2} mv^2 \)

### Step 3: Set up the equations for both cases
1. For the first case (with \( \lambda_1 = 480 \) nm):
\[ \frac{hc}{480} - \phi = \frac{1}{2} mv^2 \]
Rearranging gives:
\[ \phi = \frac{hc}{480} - \frac{1}{2} mv^2 \quad \text{(Equation 1)} \]

2. For the second case (with unknown wavelength \( \lambda \)):
\[ \frac{hc}{\lambda} - \phi = \frac{1}{2} m(2v)^2 \]
Rearranging gives:
\[ \phi = \frac{hc}{\lambda} - 2mv^2 \quad \text{(Equation 2)} \]

### Step 4: Equate the two expressions for the work function
Since \( \phi \) is the same in both cases, we can set Equation 1 equal to Equation 2:
\[ \frac{hc}{480} - \frac{1}{2} mv^2 = \frac{hc}{\lambda} - 2mv^2 \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ \frac{hc}{\lambda} = \frac{hc}{480} + \frac{3}{2} mv^2 \]

### Step 6: Isolate \( \frac{hc}{\lambda} \)
Now, we can express \( \frac{hc}{\lambda} \) in terms of \( v \):
\[ \frac{hc}{\lambda} = \frac{hc}{480} + \frac{3}{2} mv^2 \]

### Step 7: Use the relationship between the two cases
From the first case, we can express \( \frac{1}{2} mv^2 \) in terms of \( \frac{hc}{480} \):
\[ \frac{1}{2} mv^2 = \frac{hc}{480} - \phi \]

### Step 8: Substitute back into the equation
Substituting this back into the equation gives us:
\[ \frac{hc}{\lambda} = \frac{hc}{480} + 3\left(\frac{hc}{480} - \phi\right) \]

### Step 9: Solve for \( \lambda \)
To find \( \lambda \), we can simplify the equation:
\[ \frac{hc}{\lambda} = \frac{hc}{480} + 3\left(\frac{hc}{480} - \phi\right) \]
This leads to:
\[ \frac{hc}{\lambda} = \frac{4hc}{480} - 3\phi \]

### Step 10: Final calculation
Now we can find \( \lambda \) by isolating it:
\[ \lambda = \frac{hc}{\frac{4hc}{480} - 3\phi} \]

### Conclusion
After simplifying and calculating, we find that:
\[ \lambda = 300 \, \text{nm} \]

### Final Answer
The wavelength that would release electrons with maximum velocity \( 2v \) is \( 300 \, \text{nm} \).

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