In a photocell bichromatic light of wavelength `2475 Å` and `6000 Å` are incident on cathode whose work function is `4.8 eV`. If a uniform magnetic field of `3 xx 10^(-5) Tesla` exists parallel to the plate , the radius of the path describe by the photoelectron will be (mass of electron ` = 9 xx 10^(-31) kg`)

Options

A. 1 cm
B. 5 cm
C. 10 cm
D. 25 cm

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Energy of Each Wavelength
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda \) is the wavelength in meters.

**For the first wavelength \( \lambda_1 = 2475 \, \text{Å} = 2475 \times 10^{-10} \, \text{m} \):**
\[ E_1 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{2475 \times 10^{-10}} \]
Calculating this gives:
\[ E_1 \approx 5.0 \, \text{eV} \]

**For the second wavelength \( \lambda_2 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \):**
\[ E_2 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{6000 \times 10^{-10}} \]
Calculating this gives:
\[ E_2 \approx 2.06 \, \text{eV} \]

### Step 2: Determine if Electrons are Ejected
The work function \( \phi \) of the cathode is given as \( 4.8 \, \text{eV} \).

- For \( E_1 = 5.0 \, \text{eV} \):
- \( E_1 > \phi \) (electron is ejected)

- For \( E_2 = 2.06 \, \text{eV} \):
- \( E_2 < \phi \) (electron is not ejected)

### Step 3: Calculate the Kinetic Energy of the Ejected Electron
The kinetic energy \( KE \) of the ejected electron from the first wavelength is:
\[ KE = E_1 - \phi = 5.0 \, \text{eV} - 4.8 \, \text{eV} = 0.2 \, \text{eV} \]

### Step 4: Convert Kinetic Energy to Joules
To convert the kinetic energy from eV to Joules:
\[ KE = 0.2 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 3.2 \times 10^{-20} \, \text{J} \]

### Step 5: Calculate the Radius of the Path in the Magnetic Field
The radius \( r \) of the circular path of the electron in a magnetic field is given by:
\[ r = \frac{mv}{qB} \]
where:
- \( m = 9 \times 10^{-31} \, \text{kg} \) (mass of electron)
- \( v = \sqrt{\frac{2KE}{m}} \) (velocity of the electron)
- \( q = 1.6 \times 10^{-19} \, \text{C} \) (charge of electron)
- \( B = 3 \times 10^{-5} \, \text{T} \) (magnetic field strength)

**Step 5.1: Calculate the Velocity \( v \)**
\[ v = \sqrt{\frac{2 \times 3.2 \times 10^{-20}}{9 \times 10^{-31}}} \]
Calculating this gives:
\[ v \approx 2.65 \times 10^5 \, \text{m/s} \]

**Step 5.2: Calculate the Radius \( r \)**
Now substituting \( v \) into the radius formula:
\[ r = \frac{(9 \times 10^{-31})(2.65 \times 10^5)}{(1.6 \times 10^{-19})(3 \times 10^{-5})} \]
Calculating this gives:
\[ r \approx 0.05 \, \text{m} = 5 \, \text{cm} \]

### Final Answer
The radius of the path described by the photoelectron is approximately **5 cm**.

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