Light of wavelength 330nm falling on a piece of metal ejects electrons with sufficient energy with required voltage `V_0` to prevent them reaching a collector. In the same set up, light of wavelength 220 nm ejects electrons which require twice the voltage `V_0` to stop them in reaching a colleator. the numerical value of voltage `V_0` is

Options

A. `(16)/(15) V`
B. `(15)/(16) V`
C. `(15)/(8) V`
D. `(8)/(15) V`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the photoelectric effect
The photoelectric effect states that when light of a certain frequency (or wavelength) hits a metal surface, it can eject electrons from that surface. The energy of the incoming photons must be greater than the work function of the metal for electrons to be emitted.

### Step 2: Write the energy equations for the two wavelengths
For the first wavelength (330 nm), the energy of the photons can be expressed as:
\[ E_1 = \frac{hc}{\lambda_1} \]
Where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \))
- \( \lambda_1 = 330 \, \text{nm} = 330 \times 10^{-9} \, \text{m} \)

For the second wavelength (220 nm), the energy of the photons is:
\[ E_2 = \frac{hc}{\lambda_2} \]
Where \( \lambda_2 = 220 \, \text{nm} = 220 \times 10^{-9} \, \text{m} \)

### Step 3: Relate the energies to the stopping voltages
The maximum kinetic energy of the emitted electrons can be expressed in terms of the stopping voltage \( V_0 \) as:
\[ KE = eV_0 \]
Where \( e \) is the charge of an electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

For the first case (330 nm):
\[ eV_0 = E_1 - \phi \]
Where \( \phi \) is the work function of the metal.

For the second case (220 nm), since it requires twice the voltage to stop the electrons:
\[ 2eV_0 = E_2 - \phi \]

### Step 4: Set up the equations
From the above, we have two equations:
1. \( eV_0 = \frac{hc}{330 \times 10^{-9}} - \phi \) (Equation 1)
2. \( 2eV_0 = \frac{hc}{220 \times 10^{-9}} - \phi \) (Equation 2)

### Step 5: Subtract the equations
Subtract Equation 1 from Equation 2:
\[ 2eV_0 - eV_0 = \left(\frac{hc}{220 \times 10^{-9}} - \phi\right) - \left(\frac{hc}{330 \times 10^{-9}} - \phi\right) \]
This simplifies to:
\[ eV_0 = \frac{hc}{220 \times 10^{-9}} - \frac{hc}{330 \times 10^{-9}} \]

### Step 6: Factor out \( hc \)
\[ eV_0 = hc \left(\frac{1}{220 \times 10^{-9}} - \frac{1}{330 \times 10^{-9}}\right) \]

### Step 7: Calculate the difference
Finding a common denominator:
\[ \frac{1}{220} - \frac{1}{330} = \frac{330 - 220}{220 \times 330} = \frac{110}{220 \times 330} = \frac{1}{2 \times 330} \]

### Step 8: Substitute values
Now substituting \( h \) and \( c \):
\[ eV_0 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2 \times 330 \times 10^{-9}} \]

### Step 9: Solve for \( V_0 \)
Calculate the right-hand side:
\[ eV_0 = \frac{1.9878 \times 10^{-25}}{6.6 \times 10^{-7}} \]
\[ eV_0 = 3.01 \times 10^{-19} \]

Now divide by the charge of the electron:
\[ V_0 = \frac{3.01 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.88 \, \text{V} \]

### Final Result
The numerical value of \( V_0 \) is approximately \( 1.88 \, \text{V} \).