To determine the minimum value of the magnetic field \( B \) that results in zero current registered by the ammeter in the photoelectric effect experiment, we need to analyze the conditions under which the emitted photoelectrons are unable to reach the collector plate. This occurs when the magnetic force acting on the electrons is equal to the electric force that would otherwise pull them toward the collector. Let's break this down step by step.
Understanding the Photoelectric Effect
The photoelectric effect occurs when light of sufficient energy strikes a material, causing the emission of electrons. The energy of the incident photons can be calculated using the equation:
E = \frac{hc}{\lambda}
where:
- E is the energy of the photon in electron volts (eV),
- h is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- c is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength of the light in meters.
Calculating Photon Energy
Given the work function \( \phi \) of the emitter is \( 2.39 \, \text{eV} \), we need to find the energy of the incident light for wavelengths between 400 nm and 600 nm. Let's calculate the energy for both extremes:
For \( \lambda = 400 \, \text{nm} = 400 \times 10^{-9} \, \text{m} \):
E_{400} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{400 \times 10^{-9}} \approx 3.1 \, \text{eV}
For \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \):
E_{600} = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{600 \times 10^{-9}} \approx 2.07 \, \text{eV}
Since the work function is \( 2.39 \, \text{eV} \), only photons with wavelengths less than 520 nm (where \( E \) exceeds \( 2.39 \, \text{eV} \)) will cause photoemission.
Conditions for Zero Current
For the current to be zero, the electrons emitted must be deflected by the magnetic field such that they do not reach the collector. The magnetic force \( F_B \) acting on a charged particle moving in a magnetic field is given by:
F_B = qvB
where:
- q is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- v is the velocity of the emitted electrons,
- B is the magnetic field strength.
The electric force \( F_E \) that would pull the electrons toward the collector is given by:
F_E = qE
where \( E \) is the electric field strength between the plates. The condition for zero current is when \( F_B = F_E \).
Finding the Velocity of the Electrons
The kinetic energy of the emitted electrons can be expressed as:
\frac{1}{2}mv^2 = E - \phi
where \( m \) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)). For the maximum energy (using \( E_{400} \)):
\frac{1}{2}mv^2 = 3.1 \, \text{eV} - 2.39 \, \text{eV} = 0.71 \, \text{eV}
Converting \( 0.71 \, \text{eV} \) to joules:
0.71 \, \text{eV} = 0.71 \times 1.6 \times 10^{-19} \, \text{J} \approx 1.136 \times 10^{-19} \, \text{J}
Now, solving for \( v \):
v = \sqrt{\frac{2 \times 1.136 \times 10^{-19}}{9.11 \times 10^{-31}}} \approx 1.63 \times 10^6 \, \text{m/s}
Equating Forces
Setting the magnetic force equal to the electric force:
qvB = qE
We can cancel \( q \) from both sides (assuming \( q \neq 0 \)):
vB = E
To find \( E \), we can use the relationship between voltage \( V \) and electric field \( E \):
E = \frac{V}{d}
where \( d = 0.1 \, \text{m} \) (the separation between the plates). Assuming a potential difference \( V \) is applied, we can express \( B \) as:
B = \frac{E}{v} = \frac{V/d}{v}
Final Calculation
To find the minimum value of \( B \), we need to determine \( V \). The maximum electric field strength \( E \) can be calculated based on the work function and the energy of the emitted electrons. However, without a specific voltage provided, we can express \( B \) in terms of \( V \):
B = \frac{V}{0.1 \times v}
Substituting \( v \):
B = \frac{V
