What is the strength of transverse magnetic field required to bend all the photoelectrons within a circle of a radius 50 cm when light of wavelength `3800 Å` is incident on a barium emitter ? (Given that work function of barium is `2.5 eV, h=6.63xx10^(-34)js, e=1.6xx10^(19)C,m=9.1xx10^(-31)kg`

Options

A. `4.88xx10^(-4)T`
B. `32xx10^(-5)T`
C. `6.32xx10^(-6)T`
D. `7.6xx10^(-8)T`

Answer

To solve the problem, we need to find the strength of the transverse magnetic field required to bend all the photoelectrons emitted from a barium emitter within a circle of radius 50 cm when light of wavelength 3800 Å is incident on it.

### Step-by-Step Solution:

1. **Calculate the Energy of the Incident Photons**:
The energy \( E \) of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 3800 \, \text{Å} = 3800 \times 10^{-10} \, \text{m} \)

Plugging in the values:
\[ E = \frac{(6.63 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{3800 \times 10^{-10} \, \text{m}} \approx 5.21 \times 10^{-19} \, \text{J} \]

To convert this energy into electron volts (1 eV = \( 1.6 \times 10^{-19} \, \text{J} \)):
\[ E \approx \frac{5.21 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.26 \, \text{eV} \]

2. **Calculate the Maximum Kinetic Energy of the Photoelectrons**:
The maximum kinetic energy \( K_{\text{max}} \) of the emitted photoelectrons is given by:
\[ K_{\text{max}} = E - \phi \]
where \( \phi = 2.5 \, \text{eV} \) (work function of barium).
\[ K_{\text{max}} = 3.26 \, \text{eV} - 2.5 \, \text{eV} = 0.76 \, \text{eV} \]

3. **Convert Kinetic Energy to Joules**:
\[ K_{\text{max}} = 0.76 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} \approx 1.216 \times 10^{-19} \, \text{J} \]

4. **Calculate the Maximum Velocity of the Photoelectrons**:
Using the kinetic energy formula:
\[ K_{\text{max}} = \frac{1}{2} mv^2 \]
Rearranging for \( v \):
\[ v = \sqrt{\frac{2K_{\text{max}}}{m}} \]
where \( m = 9.1 \times 10^{-31} \, \text{kg} \).
\[ v = \sqrt{\frac{2 \times 1.216 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 5.0 \times 10^6 \, \text{m/s} \]

5. **Calculate the Required Magnetic Field**:
The radius \( r \) of the circular path of the electron in a magnetic field \( B \) is given by:
\[ r = \frac{mv}{eB} \]
Rearranging for \( B \):
\[ B = \frac{mv}{er} \]
where \( e = 1.6 \times 10^{-19} \, \text{C} \) and \( r = 0.5 \, \text{m} \).
\[ B = \frac{(9.1 \times 10^{-31})(5.0 \times 10^6)}{(1.6 \times 10^{-19})(0.5)} \approx 5.77 \times 10^{-6} \, \text{T} \]

### Final Answer:
The strength of the transverse magnetic field required is approximately \( 5.77 \times 10^{-6} \, \text{T} \).