Question

In an experiment on photoelectric effect, the emitter and the collector plates are placed at a separation of 10 cm and are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. The work function of the emitter is 2.39 eV and the light incident on it has wavelengths between 400 nm and 600 nm. Find the minimum value of B for which the current registered by the ammeter is zero. Neglect any effect of space charge.

Answer 1

Given:
Separation between the collector and emitter, d = 10 cm
Work function, ϕ = 2.39 eV
Wavelength range, λ₁ = 400 nm to λ₂ = 600 nm
Magnetic field B will be minimum if energy is maximum.
For maximum energy, wavelength λ should be minimum.
Einstein's photoelectric equation:
$$\begin{gathered} E=\frac{hc}{\lambda }-\varphi \\ \text{H}\mathrm{ere},h=\mathrm{Planck}\text{'s}\mathrm{constant} \\ \lambda =\mathrm{Wavelength}\mathrm{of}\mathrm{light} \\ c=\mathrm{speed}\mathrm{of}\mathrm{light} \\ \therefore E=\frac{1242}{400}-2.39 \\ =3.105-2.39=0.715\mathrm{eV} \end{gathered}$$

The beam of ejected electrons will be bent by the magnetic field. If the electrons do not reach the other plates, there will be no current.
When a charged particle is sent perpendicular to a magnetic field, it moves along a circle of radius,
$r=\frac{mv}{qB}$,
where m = mass of charge particle
B = magnetic field
v = velocity of particle
q = charge on the particle

Radius of the circle should be equal to r = d, so that no current flows in the circuit.
$$\begin{gathered} \Rightarrow r=\sqrt{\frac{2mE}{q\mathrm{B}}}\left(\because mv=\sqrt{2mE}\right) \\ \Rightarrow 0.1=\sqrt{\frac{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 0.715}{1.6\times {10}^{-19}\times B}} \\ \Rightarrow B=\sqrt{\frac{2\times 9.1\times 1.6\times 0.715\times {10}^{-50}}{1.6\times {10}^{-20}}} \\ \Rightarrow B=\sqrt{\frac{2\times 9.1\times 1.6\times 0.715}{1.6}}\times {10}^{-5} \\ \Rightarrow B=2.85\times {10}^{-5}\mathrm{T} \end{gathered}$$