A metal surface is illuminated by a radiation of wavelength 4500 $\mathop A\limits^o$. The ejected photo-electron enters a constant magnetic field of 2 mT making an angle of 90$^\circ$ with the magnetic field. If it starts revolving in a circular path of radius 2 mm, the work function of the metal is approximately :
Solution
<p>${{hc} \over \lambda } - \phi = KE$ ...... (i)</p>
<p>$R = {{mv} \over {Bq}} = {{\sqrt {2m(KE)} } \over {Bq}}$ ...... (ii)</p>
<p>Putting the values,</p>
<p>$\phi \simeq 1.36$ eV</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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