JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 14)
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: ($${{{hc} \over e}}$$ = 1240 nm eV)
0.5 V
1.0 V
2.0 V
1.5 V
Explanation
$${{hc} \over {{\lambda _1}}} = \phi + e$$V1 . . . (i)
$${{hc} \over {{\lambda _2}}} = \phi + e$$V2 . . . (ii)
(i) $$-$$ (ii)
hc$$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$$ = e(V1 $$-$$ V2)
$$ \Rightarrow $$ V1 $$-$$ V2 = $${{hc} \over e}$$$$\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} \right)$$
= (1240nm $$-$$ V) $${{100nm} \over {300nm \times 400nm}}$$
= 1V
$${{hc} \over {{\lambda _2}}} = \phi + e$$V2 . . . (ii)
(i) $$-$$ (ii)
hc$$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} \right)$$ = e(V1 $$-$$ V2)
$$ \Rightarrow $$ V1 $$-$$ V2 = $${{hc} \over e}$$$$\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} \right)$$
= (1240nm $$-$$ V) $${{100nm} \over {300nm \times 400nm}}$$
= 1V
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