JEE MAIN - Physics (2005 - No. 24)
The electrical conductivity of a semiconductor increases when electromagnetic radiation of wavelength shorter than $$2480$$ $$nm$$ is incident on it. The band gap in $$(eV)$$ for the semiconductor is
$$2.5$$ $$eV$$
$$1.1$$ $$eV$$
$$0.7$$ $$eV$$
$$0.5$$ $$eV$$
Explanation
Band gap $$=$$ energy of photon of wavelength $$2480$$ $$nm.$$ So,
$$\Delta E = {{hc} \over \lambda }$$
$$ = \left( {{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {2480 \times {{10}^{ - 9}}}}} \right) \times {1 \over {1.6 \times {{10}^{ - 19}}}}eV$$
$$ = 0.5\,eV$$
$$\Delta E = {{hc} \over \lambda }$$
$$ = \left( {{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {2480 \times {{10}^{ - 9}}}}} \right) \times {1 \over {1.6 \times {{10}^{ - 19}}}}eV$$
$$ = 0.5\,eV$$
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