JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 5)
Given figure shows few data points in a phot electric effect experiment for a certain metal. The
minimum energy for ejection of electron from its surface is:
(Plancks constant h = 6.62 × 10–34 J.s)_4th_September_Morning_Slot_en_5_1.png)
(Plancks constant h = 6.62 × 10–34 J.s)
2.10 eV
2.27 eV
2.59 eV
1.93 eV
Explanation
Kmax = hf - $$\phi $$
$$ \Rightarrow $$ eV0 = hf - $$\phi $$
at B V0 = 0, f = 5.5
$$ \therefore $$ $$\phi $$ = hf
= $${6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}}$$ J
= $${{6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}} \over {1.6 \times {{10}^{ - 19}}}}$$ eV
= 2.27 eV
$$ \Rightarrow $$ eV0 = hf - $$\phi $$
at B V0 = 0, f = 5.5
$$ \therefore $$ $$\phi $$ = hf
= $${6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}}$$ J
= $${{6.62 \times {{10}^{ - 34}} \times 5.5 \times {{10}^{14}}} \over {1.6 \times {{10}^{ - 19}}}}$$ eV
= 2.27 eV
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