JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 22)
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 :
1.8eV
0.8eV
1.1eV
1.6eV
Explanation
Let the work function be $$\phi $$.
$$ \therefore $$ KEmax = $${{hc} \over \lambda } - \phi $$
We know r = $${{mv} \over {qB}}$$
and p = mv = rqB
$$ \therefore $$ KEmax = $${{{p^2}} \over {2m}}$$ = $${{{q^2}{r^2}{B^2}} \over {2m}}$$
= $${{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {10 \times {{10}^{ - 3}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}} \over {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}$$
= 0.8 eV
$${{hc} \over \lambda }$$ = $${{12420} \over {6561}}$$ = 1.9 eV
$$ \therefore $$ $$\phi $$ = 1.9 - 0.8 = 1.1 eV
$$ \therefore $$ KEmax = $${{hc} \over \lambda } - \phi $$
We know r = $${{mv} \over {qB}}$$
and p = mv = rqB
$$ \therefore $$ KEmax = $${{{p^2}} \over {2m}}$$ = $${{{q^2}{r^2}{B^2}} \over {2m}}$$
= $${{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}{{\left( {10 \times {{10}^{ - 3}}} \right)}^2}{{\left( {3 \times {{10}^{ - 4}}} \right)}^2}} \over {2 \times 9 \times {{10}^{ - 31}} \times 1.6 \times {{10}^{ - 19}}}}$$
= 0.8 eV
$${{hc} \over \lambda }$$ = $${{12420} \over {6561}}$$ = 1.9 eV
$$ \therefore $$ $$\phi $$ = 1.9 - 0.8 = 1.1 eV
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