Which combination of factors is most suitable for increasing the electrical conductivity of metals? a. more valence electrons and larger atomic radius. b. fewer valence electrons and larger atomic radius. c. more valence electrons and smaller atomic radius. d. fewer valence electrons and smaller atomic radius

The electrical conductivity of a metal depends on the free electron concentration.A substance having more number of free electrons in its conduction band has more conductivity than others.

Free electrons are these electrons which experience less or negligible force from the nucleus of an atom.In case of metals, the valence band and conduction band are either overlapped or they have a small forbidden energy gap.Hence,the metal having large number of valence electrons will be responsible for higher conductivity of the specimen.

The atomic radius also plays a crucial role in determining the conductivity of a metal.In case of metals whose atoms have lesser atomic radius,the nucleus exerts higher amount of attractive coulombic force on electrons.Hence the electron can not be free.Larger the atomic radius,the lesser will be the force of attraction of nucleus on the electrons.Hence,the availability of free electrons will be more.

Hence the correct option will be A.more valence electrons and larger atomic radius

option a . more valence electron and larger atomic radius

Explanation:

There are two main factor that determine the strength of electrical conductivity of metals. the factors are free electron and atomic radius.

Higher the number of free electron higher will be the chances of high electrical conductivity as conductivity is dependent on the number of free electron

smaller the atomic radius, larger will be cohesion between the atomic particle hence offer low electrical conductivity.

larger the atomic radius, lesser will be the cohesion between the particle thus offer high electrical conductivity

the answer is a. at the equator

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)