Radiation from a hydrogen discharge tube ( energy of photons `le 13.6 eV`) goes through a fliter which transmits only waves of wavelength greater than `4400Å` and is incident on a metal of work function `2.0 eV`. If stopping potential is `nxx10^(-2)` volts. Find the value of `'n'`

Answer

To solve the given problem step by step, we will follow the principles of the photoelectric effect and the relationship between energy, wavelength, and stopping potential.

### Step 1: Identify the given values
- Energy of photons from the hydrogen discharge tube, \( E = 13.6 \, \text{eV} \)
- Work function of the metal, \( W = 2.0 \, \text{eV} \)
- Wavelength cutoff from the filter, \( \lambda = 4400 \, \text{Å} \)

### Step 2: Calculate the energy of the photons that can be transmitted through the filter
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
Where:
- \( h \) is Planck's constant, \( h \approx 4.1357 \times 10^{-15} \, \text{eV s} \)
- \( c \) is the speed of light, \( c \approx 3 \times 10^8 \, \text{m/s} \)
- \( \lambda \) in meters is \( 4400 \, \text{Å} = 4400 \times 10^{-10} \, \text{m} = 4.4 \times 10^{-7} \, \text{m} \)

Calculating \( E \):
\[ E = \frac{(4.1357 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{4.4 \times 10^{-7} \, \text{m}} \approx 2.83 \, \text{eV} \]

### Step 3: Determine the maximum kinetic energy of the emitted electrons
According to the photoelectric effect, the maximum kinetic energy \( K.E. \) of the emitted electrons is given by:
\[ K.E. = E - W \]
Substituting the values:
\[ K.E. = 2.83 \, \text{eV} - 2.0 \, \text{eV} = 0.83 \, \text{eV} \]

### Step 4: Relate the kinetic energy to the stopping potential
The stopping potential \( V_0 \) is related to the maximum kinetic energy by:
\[ K.E. = eV_0 \]
Where \( e \) is the charge of an electron (1 eV corresponds to the energy gained by an electron when accelerated through a potential difference of 1 volt). Thus:
\[ V_0 = K.E. \]
So:
\[ V_0 = 0.83 \, \text{V} \]

### Step 5: Express the stopping potential in terms of \( n \)
Given that the stopping potential is expressed as \( V_0 = n \times 10^{-2} \, \text{V} \), we can equate:
\[ 0.83 = n \times 10^{-2} \]

### Step 6: Solve for \( n \)
Rearranging gives:
\[ n = 0.83 \times 10^{2} = 83 \]

### Final Answer
The value of \( n \) is \( 83 \).

---