JEE MAIN - Chemistry (2018 - 15th April Morning Slot - No. 21)
Ejection of the photoelectron from metal in the photoelectric experiment can be stopped by applying 0.5 V when the radiation of 250 nm is used. The work function of the metal is :
4 eV
4.5 eV
5 eV
5.5 eV
Explanation
$$\lambda $$ = 250 nm = 2500 $$\mathop A\limits^ \circ $$
E = $${{hc} \over \lambda }$$ = $${{12400} \over {2500}}$$ = 4.96 eV
K. E = 0.5 eV
As, $$\,\,\,\,\,\,$$ K. E = E $$-$$ $$\omega $$0
$$ \Rightarrow $$ $$\,\,\,$$ 0.5 = 4.96 $$-$$ $${\omega _0}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${\omega _0}$$ = 4.46 eV $$ \simeq $$ 4.5 eV
E = $${{hc} \over \lambda }$$ = $${{12400} \over {2500}}$$ = 4.96 eV
K. E = 0.5 eV
As, $$\,\,\,\,\,\,$$ K. E = E $$-$$ $$\omega $$0
$$ \Rightarrow $$ $$\,\,\,$$ 0.5 = 4.96 $$-$$ $${\omega _0}$$
$$ \Rightarrow $$ $$\,\,\,$$ $${\omega _0}$$ = 4.46 eV $$ \simeq $$ 4.5 eV
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