JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 19)
If a semiconductor photodiode can detect a photon with a maximum wavelength of 400 nm, then
its band gap energy is :
Planck’s constant h = 6.63 $$ \times $$ 10–34 J.s. Speed of light c = 3 $$ \times $$ 108 m/s
Planck’s constant h = 6.63 $$ \times $$ 10–34 J.s. Speed of light c = 3 $$ \times $$ 108 m/s
1.5 eV
2.0 eV
3.1 eV
1.1 eV
Explanation
$$E = {{hc} \over \lambda }$$
= $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}$$
= $${{1240} \over {400}}$$ eV
= 3.1 eV
= $${{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {400 \times {{10}^{ - 9}}}}$$
= $${{1240} \over {400}}$$ eV
= 3.1 eV
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