Radiation, with wavelength 6561 $\mathop A\limits^o$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of 3 × 10–4 T. If the radius of the largest circular path followed by the electrons is 10 mm, the work function of the metal is close to :
Solution
Let the work function be $\phi$.
<br><br>$\therefore$ KE<sub>max</sub> = ${{hc} \over \lambda } - \phi$
<br><br>We know r = ${{mv} \over {qB}}$
<br><br>and p = mv = rqB
<br><br>$\therefore$ KE<sub>max</sub> = ${{{p^2}} \over {2m}}$ = ${{{q^2}{r^2}{B^2}} \over {2m}}$
<br><br> = $${{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {10 \times {{10}^{ - 3}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}} \over {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}$$
<br><br>= 0.8 eV
<br><br>${{hc} \over \lambda }$ = ${{12420} \over {6561}}$ = 1.9 eV
<br><br>$\therefore$ $\phi$ = 1.9 - 0.8 = 1.1 eV
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
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