To solve the problem step by step, we will follow the principles of the photoelectric effect and the motion of charged particles in a magnetic field.
### Step 1: Calculate the energy of the incident photons
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 two wavelengths given:
1. For \(\lambda_1 = 4000 \, \text{Å} = 4000 \times 10^{-10} \, \text{m}\):
\[
E_1 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{4000 \times 10^{-10}} \approx 3.1 \, \text{eV}
\]
2. For \(\lambda_2 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m}\):
\[
E_2 = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{6000 \times 10^{-10}} \approx 2.06 \, \text{eV}
\]
### Step 2: Determine if the photons can cause photoemission
The work function \(\phi\) is given as \(2.39 \, \text{eV}\). We compare the energies of the photons with the work function:
- \(E_1 = 3.1 \, \text{eV} > \phi\) (photoemission occurs)
- \(E_2 = 2.06 \, \text{eV} < \phi\) (no photoemission)
Only photons of wavelength \(4000 \, \text{Å}\) will cause photoemission.
### Step 3: Calculate the kinetic energy of the emitted electrons
The kinetic energy \(K\) of the emitted electrons can be calculated as:
\[
K = E_1 - \phi = 3.1 \, \text{eV} - 2.39 \, \text{eV} = 0.71 \, \text{eV}
\]
### Step 4: Convert kinetic energy to joules
To work with SI units, we convert the kinetic energy from electron volts to joules:
\[
K = 0.71 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 1.136 \times 10^{-19} \, \text{J}
\]
### Step 5: Determine the radius of the semicircular path
The radius \(R\) of the semicircular path of the electrons in the magnetic field is given as \(R = 0.05 \, \text{m}\) (5 cm).
### Step 6: Apply the formula for the radius of the path in a magnetic field
The radius of the path of a charged particle in a magnetic field is given by:
\[
R = \frac{mv}{Bq}
\]
where:
- \(m\) is the mass of the electron \(m = 9.109 \times 10^{-31} \, \text{kg}\)
- \(v\) is the velocity of the electron
- \(B\) is the magnetic field strength
- \(q\) is the charge of the electron \(q = 1.6 \times 10^{-19} \, \text{C}\)
### Step 7: Relate kinetic energy to velocity
The kinetic energy is also related to the velocity as:
\[
K = \frac{1}{2} mv^2 \implies v = \sqrt{\frac{2K}{m}}
\]
Substituting the value of \(K\):
\[
v = \sqrt{\frac{2 \times 1.136 \times 10^{-19}}{9.109 \times 10^{-31}}} \approx 1.63 \times 10^6 \, \text{m/s}
\]
### Step 8: Substitute into the radius formula to find \(B\)
Rearranging the radius formula to solve for \(B\):
\[
B = \frac{mv}{Rq}
\]
Substituting the values:
\[
B = \frac{(9.109 \times 10^{-31})(1.63 \times 10^6)}{(0.05)(1.6 \times 10^{-19})}
\]
Calculating:
\[
B \approx 5.68 \times 10^{-5} \, \text{T}
\]
### Final Answer
The minimum value of the magnetic field \(B\) for which the galvanometer shows zero deflection is approximately:
\[
B \approx 5.68 \times 10^{-5} \, \text{T}
\]
