A beam of 450 nm light is incident on a metal having work function 2eV and placed in a magnetic field B. If the most energetic electrons emitted are bent into circular are of radius 0.2 m, find B.

Options

A. `2.36 xx 10^(-4)`T
B. `1.46 xx 10^(-5)` T
C. `6.9 xx 10^(-5)T`
D. `9.2 xx 10^(-6)T`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the energy of the incident light
The energy of a photon can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(h\) = Planck's constant = \(4.14 \times 10^{-15} \, \text{eV s}\)
- \(c\) = speed of light = \(3 \times 10^8 \, \text{m/s}\)
- \(\lambda\) = wavelength of light = \(450 \, \text{nm} = 450 \times 10^{-9} \, \text{m}\)

Substituting the values:
\[ E = \frac{(4.14 \times 10^{-15} \, \text{eV s})(3 \times 10^8 \, \text{m/s})}{450 \times 10^{-9} \, \text{m}} = 2.76 \, \text{eV} \]

### Step 2: Calculate the maximum kinetic energy of the emitted electrons
Using Einstein's photoelectric equation:
\[ K_{\text{max}} = E - \phi \]
where \(\phi\) is the work function of the metal (given as \(2 \, \text{eV}\)):
\[ K_{\text{max}} = 2.76 \, \text{eV} - 2 \, \text{eV} = 0.76 \, \text{eV} \]

### Step 3: Convert kinetic energy to joules
To convert the kinetic energy from eV to joules, we use:
\[ K_{\text{max}} = 0.76 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.216 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the maximum velocity of the emitted electrons
Using the kinetic energy formula:
\[ K_{\text{max}} = \frac{1}{2} mv^2 \]
Rearranging for \(v\):
\[ v = \sqrt{\frac{2K_{\text{max}}}{m}} \]
where \(m\) is the mass of an electron (\(9.1 \times 10^{-31} \, \text{kg}\)):
\[ v = \sqrt{\frac{2 \times 1.216 \times 10^{-19} \, \text{J}}{9.1 \times 10^{-31} \, \text{kg}}} = 0.52 \times 10^6 \, \text{m/s} \]

### Step 5: Calculate the magnetic field
The radius \(r\) of the circular path of the electrons in a magnetic field is given by:
\[ r = \frac{mv}{eB} \]
Rearranging for \(B\):
\[ B = \frac{mv}{er} \]
Substituting the known values:
- \(e = 1.6 \times 10^{-19} \, \text{C}\)
- \(r = 0.2 \, \text{m}\)

\[ B = \frac{(9.1 \times 10^{-31} \, \text{kg})(0.52 \times 10^6 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C})(0.2 \, \text{m})} \]

Calculating \(B\):
\[ B = \frac{4.732 \times 10^{-25}}{3.2 \times 10^{-20}} = 1.48 \times 10^{-5} \, \text{T} \]

### Final Answer
Thus, the magnetic field \(B\) is approximately:
\[ B \approx 1.46 \times 10^{-5} \, \text{T} \]