JEE MAIN - Physics (2014 (Offline) - No. 4)
The radiation corresponding to $$3 \to 2$$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field $$3 \times {10^{ - 4}}\,T.$$ If the radius of the larger circular path followed by these electrons is $$10.0$$ $$mm$$, the work function of the metal is close to:
$$1.8$$ $$eV$$
$$1.1$$ $$eV$$
$$0.8$$ $$eV$$
$$1.6$$ $$eV$$
Explanation
Radius of circular path followed by electron is given by,
$$r = {{m\upsilon } \over {qB}} = {{\sqrt {2meV} } \over {eB}} = {1 \over B}\sqrt {{{2m} \over e}V} $$
$$ \Rightarrow V = {{{B^2}{r^2}e} \over {2m}} = 0.8V$$
For transition between $$3$$ to $$2.$$
$$E = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$$
$$ = {{13.6 \times 5} \over {36}} = 1.88eV$$
Work function $$ = 1.88eV - 0.8eV$$
$$ = 1.08eV \approx 1.1eV$$
$$r = {{m\upsilon } \over {qB}} = {{\sqrt {2meV} } \over {eB}} = {1 \over B}\sqrt {{{2m} \over e}V} $$
$$ \Rightarrow V = {{{B^2}{r^2}e} \over {2m}} = 0.8V$$
For transition between $$3$$ to $$2.$$
$$E = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$$
$$ = {{13.6 \times 5} \over {36}} = 1.88eV$$
Work function $$ = 1.88eV - 0.8eV$$
$$ = 1.08eV \approx 1.1eV$$
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