Question

In an experiment on photoelectric effect, the separation between emitter and the collector plates is $10cm$. Plates are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. The work function of the emitter is $2.39eV$ and the light of wavelengths between $400nm$ and $600nm$ is incident on it. If minimum value of $B$ for which the current registered by the ammeter is zero is $n\times {10}^{-6}T$. Then find out value of $n$ (Neglect any effect of space charge). (assume emission of photo electron to be radomly every possible direction)

Answer 1

Here; ${\Phi }_0=2.39eV$

${\lambda }_1=400nm$ and ${\lambda }_2=600nm$

For ${}^{\prime }{B}^{\prime }$ to be minimum Energy should be Maximum

$\therefore$ $\lambda$ sgould be minimum

Here; $E=\frac{hc}{\lambda }-{\Phi }_0$

$=3.015-2.39$

$=0.715eV$

The presence of M.field will bend the beam there will be no current if the electron does'not reach the other plates.

$\therefore r=\frac{mv}{qB}\Rightarrow r=\frac{\sqrt{2mE}}{qB}$

$\Rightarrow B=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 0.715}}{1.6\times {10}^{-19}\times 0.1}$

$=2.85\times {10}^{-5}$

$=0.285\times {10}^{-6}Tesla$