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In simple metals the phonon contribution to the electrical resistivity at temperature T is

A. directly proportional to T above Debye temperature and to T3 below it

B. inversely proportional to T for all temperatures

C. independent of T for all temperatures

D. directly proportional to T above Debye temperature and to T5 below it

Answer: Option A

Solution (By Examveda Team)

In simple metals, the phonon contribution to the electrical resistivity at temperature T follows specific trends based on the Debye temperature:

Option A: This option is correct. Above the Debye temperature, the phonon contribution to the electrical resistivity is directly proportional to T. Below the Debye temperature, it is proportional to T3. This behavior is in accordance with the Debye theory of solids, which predicts these dependencies.

Option B: This option is incorrect. The phonon contribution to electrical resistivity is not inversely proportional to T for all temperatures in simple metals.

Option C: This option is incorrect. The phonon contribution to electrical resistivity is not independent of temperature for all temperatures in simple metals.

Option D: This option is incorrect. While it states a direct proportionality above the Debye temperature, it incorrectly suggests a proportionality to T5 below the Debye temperature. The correct dependency below the Debye temperature is T3, not T5. Therefore, this option is not accurate.

This Question Belongs to Engineering Physics >> Solid State Physics

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Comments (1)

  1. Manoj Kumar
    Manoj Kumar:
    1 year ago

    How you get this proportionality relation?

Related Questions on Solid State Physics

In a cubic crystal, atoms of mass M1 lie on one set of planes and atoms of mass M2 lie on planes interleaved between those of the first set. If C is the forte constant between nearest neighbour planes, the frequency of lattice vibrations for the optical phonon branch with wave vector k = 0 is

A. $$\sqrt {2C\left( {\frac{1}{{{M_1}}} + \frac{1}{{{M_2}}}} \right)} $$

B. $$\sqrt {C\left( {\frac{1}{{2{M_1}}} + \frac{1}{{{M_2}}}} \right)} $$

C. $$\sqrt {C\left( {\frac{1}{{{M_1}}} + \frac{1}{{2{M_2}}}} \right)} $$

D. zero