A) \[{{({{E}_{g}})}_{C}}={{({{E}_{g}})}_{Si}}={{({{E}_{g}})}_{Ge}}\]
B) \[{{({{E}_{g}})}_{C}}>{{({{E}_{g}})}_{Si}}>{{({{E}_{g}})}_{Ge}}\]
C) \[{{({{E}_{g}})}_{C}}<{{({{E}_{g}})}_{Ge}}>{{({{E}_{g}})}_{Si}}\]
D) \[{{({{E}_{g}})}_{Si}}<{{({{E}_{g}})}_{Ge}}>{{({{E}_{g}})}_{C}}\]
Correct Answer: B
Solution :
Carbon [C], germanium (Ge) and silicon (Si) are semiconductors. They lie in IV A group of periodic table. Now, \[{{({{E}_{g}})}_{C}}=5.2\,eV\] \[{{({{E}_{g}})}_{Ge}}=0.75\,eV\] \[{{({{E}_{g}})}_{Si}}=1.21\,eV\] Thus, it is obvious that \[{{({{E}_{g}})}_{C}}>{{({{E}_{g}})}_{Si}}>{{({{E}_{g}})}_{Ge}}\]You need to login to perform this action.
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