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प्रश्न
Pieces of copper and of silicon are initially at room temperature. Both are heated to temperature T. The conductivity of ______.
विकल्प
both increases.
both decreases.
copper increases and silicon decreases.
copper decreases and silicon increases.
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उत्तर
Pieces of copper and of silicon are initially at room temperature. Both are heated to temperature T. The conductivity of copper decreases and silicon increases.
Explanation:
The resistance of copper wire increases with temperature, which causes a drop in conductivity, while the resistance of silicon wire reduces with temperature, which causes an increase in conductivity.
संबंधित प्रश्न
In a p-type semiconductor, the acceptor impurity produces an energy level ______
A donor impurity results in ______.
Semiconductors formed by doping germanium (Ge) with aluminium (Z = 13) and antimony (Z = 51) are ______.
In n-type semiconductor majority carriers and minority carriers are respectively ______.
In p-type semiconductor, the dopant is ______.
Distinguish between n-type and p-type semiconductors.
Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV?
Suppose a ‘n’-type wafer is created by doping Si crystal having 5 × 1028 atoms/m3 with 1 ppm concentration of As. On the surface 200 ppm Boron is added to create ‘P’ region in this wafer. Considering n i = 1.5 × 1016 m–3, (i) Calculate the densities of the charge carriers in the n and p regions. (ii) Comment which charge carriers would contribute largely for the reverse saturation current when diode is reverse biased.
The figure shows a piece of pure semiconductor S in series with a variable resistor R and a source of constant voltage V. Should the value of R be increased or decreased to keep the reading of the ammeter constant, when semiconductor S is heated? Justify your answer
- Assertion (A): Putting the p-type semiconductor slab directly in physical contact with the n-type semiconductor slab cannot form the pn junction.
- Reason (R): The roughness at contact will be much more than inter atomic crystal spacing and continuous flow of charge carriers is not possible.
