When barium is irradiated by a light of `lamda=4000A` all the photoelectrons emitted are bent in a circle of radius 50 cm by a magnetic field of flux density `5.26xx10^(-6)` T acting perpendicular to plane of emission of photoelectron. Then,

Options

A. the kinetic energy of fastest photoelectrin is 0.6 eV
B. work function of the metal is 2.5 eV
C. the maximum velocity of photoelectron is `0.46xx10^(6)ms^(-1)`
D. the stopping potential for photoelectric effect is 0.6 V

Explanation

To solve the problem step by step, we will follow the logical sequence of calculations based on the information provided.

### Step 1: Calculate the Velocity of Photoelectrons
The formula for the velocity \( V \) of the photoelectrons in a magnetic field is given by:

\[ V = \frac{B \cdot q \cdot R}{m} \]

Where:
- \( B = 5.26 \times 10^{-6} \, \text{T} \) (magnetic flux density)
- \( q = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron)
- \( R = 0.5 \, \text{m} \) (radius of the circular path)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \) (mass of an electron)

Substituting the values into the formula:

\[ V = \frac{(5.26 \times 10^{-6}) \cdot (1.6 \times 10^{-19}) \cdot (0.5)}{9.1 \times 10^{-31}} \]

Calculating this gives:

\[ V \approx 0.46 \times 10^{6} \, \text{m/s} \]

### Step 2: Calculate the Kinetic Energy of the Photoelectrons
The kinetic energy \( K \) of the photoelectrons is given by:

\[ K = \frac{1}{2} m V^2 \]

Substituting the values:

\[ K = \frac{1}{2} \cdot (9.1 \times 10^{-31}) \cdot (0.46 \times 10^{6})^2 \]

Calculating this gives:

\[ K \approx 0.962 \times 10^{-19} \, \text{J} \]

### Step 3: Convert Kinetic Energy to Electron Volts
To convert the kinetic energy from joules to electron volts, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):

\[ K_e = \frac{0.962 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 0.6 \, \text{eV} \]

### Step 4: Calculate the Stopping Potential
The stopping potential \( V_s \) is equal to the kinetic energy in electron volts:

\[ V_s = 0.6 \, \text{V} \]

### Step 5: Calculate the Work Function
The work function \( W \) can be calculated using the formula:

\[ K_e = \frac{hc}{\lambda} - W \]

Where:
- \( h = 4.14 \times 10^{-15} \, \text{eV s} \)
- \( c = 3 \times 10^{8} \, \text{m/s} \)
- \( \lambda = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m} \)

Calculating \( \frac{hc}{\lambda} \):

\[ \frac{hc}{\lambda} = \frac{(4.14 \times 10^{-15}) \cdot (3 \times 10^{8})}{4000 \times 10^{-10}} \approx 2.5 \, \text{eV} \]

Now substituting back to find \( W \):

\[ 0.6 = 2.5 - W \implies W = 2.5 - 0.6 = 1.9 \, \text{eV} \]

### Summary of Results
- Velocity of photoelectrons: \( 0.46 \times 10^{6} \, \text{m/s} \)
- Kinetic Energy: \( 0.6 \, \text{eV} \)
- Stopping Potential: \( 0.6 \, \text{V} \)
- Work Function: \( 1.9 \, \text{eV} \)