
Let \[{\sigma _1}\] and \[{\sigma _2}\] are the electrical conductivities of \[Ge\] and \[Na\;\] respectively. If these substance are heated, then
A. Both \[{\sigma _1}\] and \[{\sigma _2}\] increase
B. \[{\sigma _1}\] increases and \[{\sigma _2}\] decreases
C. \[{\sigma _1}\] decreases and \[{\sigma _2}\] increases
D. Both \[{\sigma _1}\] and \[{\sigma _2}\] decrease
Answer
163.8k+ views
Hint:We know that the conductivity varies differently for metals and semiconductors with the change in temperature. So, the conductivities of \[Ge\] and \[Na\;\] vary differently with change in temperature. As temperature rises, semiconductor conductivity rises while metal conductivity falls.
Formula used:
Conductivity, $\sigma = \dfrac{{n{e^2}\tau }}{m}$
where \[n\] is the number density of charge carriers, \[e\] is the charge on each carrier, $\tau $ is the relaxation time and \[m\] is the mass of charge carriers.
Complete step by step solution:
Since, conductivity of metals and semiconductors vary differently with the change in temperature, therefore, the conductivities of \[Ge\] and \[Na\;\] will vary differently with increase or decrease in temperature. This is so because \[Ge\] is a semiconductor whereas \[Na\;\] is a metal.
As temperature rises, more electrons collide with fixed lattice ions and atoms, which cause relaxation time $\left( \tau \right)$ to shorten. As a result, metal conductivity declines as temperature increases. This can be deduced from the relationship between the conductivity and the relaxation time. The relationship so used is,
$\sigma = \dfrac{{n{e^2}\tau }}{m}$
Since charge carriers (electrons and holes) become free as the temperature rises in the case of semiconductors, \[n\] increases, conductivity rises along with temperature.
Hence, option B is the correct answer.
Note: In such types of questions different types of conductors, like metallic conductors, ionic conductors, semiconductors etc., may be asked to compare. The conductivity of an ionic conductor increases as temperature rises because this causes the release of positive and negative ions, which serve as charge carriers in ionic conductors.
Formula used:
Conductivity, $\sigma = \dfrac{{n{e^2}\tau }}{m}$
where \[n\] is the number density of charge carriers, \[e\] is the charge on each carrier, $\tau $ is the relaxation time and \[m\] is the mass of charge carriers.
Complete step by step solution:
Since, conductivity of metals and semiconductors vary differently with the change in temperature, therefore, the conductivities of \[Ge\] and \[Na\;\] will vary differently with increase or decrease in temperature. This is so because \[Ge\] is a semiconductor whereas \[Na\;\] is a metal.
As temperature rises, more electrons collide with fixed lattice ions and atoms, which cause relaxation time $\left( \tau \right)$ to shorten. As a result, metal conductivity declines as temperature increases. This can be deduced from the relationship between the conductivity and the relaxation time. The relationship so used is,
$\sigma = \dfrac{{n{e^2}\tau }}{m}$
Since charge carriers (electrons and holes) become free as the temperature rises in the case of semiconductors, \[n\] increases, conductivity rises along with temperature.
Hence, option B is the correct answer.
Note: In such types of questions different types of conductors, like metallic conductors, ionic conductors, semiconductors etc., may be asked to compare. The conductivity of an ionic conductor increases as temperature rises because this causes the release of positive and negative ions, which serve as charge carriers in ionic conductors.
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