Two metallic plates P (collector) and Q (emitter) are separated by a distance of 0.1 m. These are connected through an ammeter without any cell. A magnetic field $\overrightarrow{\mathrm{B}}$ exists parallel to the plates. Light of wavelengths between 4000 Å and 6000 Å fall on the plate Q whose work function is 2.39 eV. Calculate the minimum value of $\overrightarrow{\mathrm{B}}$ for which the current registered with ammeter is zero.

Given Data :

Two metallic plates P (collector) and Q (emitter) are separated by a distance of $0.1m$. These are connected through an ammeter without any cell. A magnetic field $\overrightarrow{B}$ exists parallel to the plates. Light of wavelengths between $4000\stackrel{0}{A}$ and$6000\stackrel{0}{A}$ fall on the plate Q whose work function is $2.39eV$

Formula Used $E=\frac{hc}{\lambda },K_{max}=\frac{1}{2}m{v}_{max}^2$

Detailed Solution:

Energy of the incident photon of light of wave length $6000\stackrel{0}{A}$ is

$E=\frac{hc}{\lambda }=\frac{\left(6.6\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{6000\times 1.6\times {10}^{-19}}\mathrm{eV}=2.01\mathrm{eV}$

$$\begin{gathered} \mathrm{Since} \mathrm{this} \mathrm{value} \mathrm{of} \mathrm{energy} \mathrm{is} \mathrm{less} \mathrm{than} \mathrm{work} \mathrm{function} ( = 2 . 39 \mathrm{eV} ) \\ \mathrm{of} \mathrm{the} \mathrm{metal}, \mathrm{hence} \mathrm{no} \mathrm{photoelectron} \mathrm{is} \mathrm{emitted} \mathrm{for} \mathrm{light} \mathrm{of} \mathrm{wavelength} \\ 6000 \stackrel{0}{\mathrm{A}} . \\ \mathrm{Since} \mathrm{the} \mathrm{energy} \mathrm{of} \mathrm{photon} \mathrm{of} \mathrm{light} \mathrm{of} \mathrm{wavelength} 4000 \stackrel{0}{\mathrm{A}} \mathrm{is} \mathrm{greater} \\ \mathrm{than} 2.39 \mathrm{eV} , \mathrm{so} \mathrm{photoelectric} \mathrm{emission} \mathrm{takes} \mathrm{place} \mathrm{with} \mathrm{its} \mathrm{halt}. \\ \mathrm{Maximum} \mathrm{K}.\mathrm{E}. \mathrm{of} \mathrm{the} \mathrm{emitted} \mathrm{photoelectron} \mathrm{is} \end{gathered}$$

$\begin{array}{r}{K}_{max}=\frac{1}{2}m{v}_{max}^2=\frac{hc}{\lambda }-{\varphi }_0\\ =\frac{\left(6.6\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{4000\times {10}^{-10}\times 1.6\times {10}^{-19}}-2.39\mathrm{eV}\\ =3.1-2.39=0.71\mathrm{eV}=0.71\times 1.6\times {10}^{-19}\mathrm{J}\\ v_{max}={\left[\frac{2\times 0.71\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\right]}^{1/2}=5\times {10}^5\mathrm{m}/\mathrm{s}\end{array}$

For Zero current

$\begin{array}{r}\frac{{\mathrm{mv}}^2}{\mathrm{r}}=\mathrm{evB}\text{ or }\mathrm{B}=\frac{\mathrm{mv}}{\mathrm{er}}\\ =\frac{\left(9.1\times {10}^{-31}\right)\times 5\times {10}^5}{\left(1.6\times {10}^{-19}\right)\times 0.1}=2.86\times {10}^{-5}\mathrm{T}\end{array}$