Question

In a photoelectric cell, the collector and emitter plates are separated by a distance of $15\text{cm}$. The photocell is connected with an ammeter in series. When white light [$\lambda$ (wavelength) between $400\text{nm}-700\text{nm}$] falls on this plate (work function $=2.1\text{eV}$), a current flows through the ammeter. A magnetic field is now switched on between the plates as shown in the figure. As this field is increased, the current through the ammeter decreases and finally becomes zero. The minimum magnetic field, for which the ammeter current becomes zero, is found to be (0.45$\times {10}^{-5}b$ Tesla, where $‘b^{\prime }$ is close to an integer. Find the value of $b$. Assume that all the photoelectrons are emitted normally from the emitter.

Answer 1

Energy of photon corresponding to wavelength $\lambda$
$E=\frac{hc}{\lambda }$
$E_{max}=\frac{hc}{{\lambda }_{min}}=\frac{(6.6\times {10}^{-34})\times (3\times {10}^8)}{4\times {10}^{-7}}\times \frac{1}{1.6\times {10}^{-19}}=3.1\text{eV}$
From Einstein's equation for photoelectric effect,
$E_{max}=\phi +(E_k{)}_{max}$
$(E_k{)}_{max}=E_{max}-\phi =3.1-2.1=1\text{eV}$

The photoelectron with charge $e$, mass $m$ and maximum velocity bends in a circle of radius $r$
$Bev=\frac{m{v}^2}{r}$
$r=\frac{mv}{Be}=\frac{\sqrt{2m{E}_k}}{Be}$

For ammeter reading to be zero, no electron should reach the collector plate.
For which $r\le d$,
$\frac{\sqrt{2m{E}_k}}{Be}\le d$
$B\ge \frac{\sqrt{2m{E}_k}}{ed}\Rightarrow B_{min}=2.25\times {10}^{-5}\text{T}$