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. 360nm
C. 384 nm
D. 300 nm

Explanation

To solve the problem step by step, we will use the principles of the photoelectric effect.

### Given:
- Wavelength of incident light, \( \lambda = 480 \, \text{nm} \)
- Cut-off wavelength, \( \lambda_0 = 600 \, \text{nm} \)
- Maximum velocity of emitted electrons when irradiated with \( \lambda = 480 \, \text{nm} \) is \( v \).
- Maximum velocity of emitted electrons when irradiated with a new wavelength \( \lambda' \) is \( 2v \).

### Step 1: Calculate the energy of the incident photons for \( \lambda = 480 \, \text{nm} \)

The energy of a photon is given by the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{Js}) \)
- \( c \) is the speed of light \( (3 \times 10^8 \, \text{m/s}) \)
- \( \lambda \) is the wavelength in meters.

Convert \( \lambda = 480 \, \text{nm} \) to meters:
\[ \lambda = 480 \times 10^{-9} \, \text{m} \]

Now calculate the energy:
\[ E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{480 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{480 \times 10^{-9}} \approx 4.14 \times 10^{-19} \, \text{J} \]

### Step 2: Calculate the work function \( W \)

The work function \( W \) can be calculated using the cut-off wavelength \( \lambda_0 = 600 \, \text{nm} \):
\[ W = \frac{hc}{\lambda_0} \]
Convert \( \lambda_0 = 600 \, \text{nm} \) to meters:
\[ \lambda_0 = 600 \times 10^{-9} \, \text{m} \]

Now calculate the work function:
\[ W = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} = \frac{1.9878 \times 10^{-25}}{600 \times 10^{-9}} \approx 3.31 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the kinetic energy of emitted electrons for \( \lambda = 480 \, \text{nm} \)

The kinetic energy \( K.E. \) of the emitted electrons can be given by:
\[ K.E. = E - W \]
Substituting the values:
\[ K.E. = 4.14 \times 10^{-19} - 3.31 \times 10^{-19} = 0.83 \times 10^{-19} \, \text{J} \]

### Step 4: Relate kinetic energy to maximum velocity

The kinetic energy is also given by:
\[ K.E. = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \, \text{kg}) \).

Setting the two expressions for kinetic energy equal:
\[ \frac{1}{2} mv^2 = 0.83 \times 10^{-19} \]

### Step 5: Calculate the new wavelength \( \lambda' \) for maximum velocity \( 2v \)

Using the same kinetic energy expression for the new maximum velocity \( 2v \):
\[ K.E. = \frac{1}{2} m (2v)^2 = 2mv^2 \]
Thus, we have:
\[ 2mv^2 = E' - W \]
Where \( E' \) is the energy of the new wavelength \( \lambda' \):
\[ E' = W + 2mv^2 \]

### Step 6: Set up the equation for the new wavelength

Using the energy relation:
\[ E' = \frac{hc}{\lambda'} \]
Substituting:
\[ \frac{hc}{\lambda'} = W + 2mv^2 \]

### Step 7: Substitute known values and solve for \( \lambda' \)

From earlier, we know \( W \) and \( mv^2 \):
\[ \lambda' = \frac{hc}{W + 2 \cdot 0.83 \times 10^{-19}} \]

After calculating, we find:
\[ \lambda' = 300 \, \text{nm} \]

### Final Answer:
\[ \lambda' = 300 \, \text{nm} \]