At 300 K, the resistivity of Be is 4 x 10-8 Ωm with a temperature coefficient ofresistivity of 0.025 K-1.a) Determine the resistivity at 400 K.b) The electrical resistivity of a beryllium alloy containing 5 at% of an alloyingelement is 50 × 10-8 Ωm at 300 K. Determine the contribution to resistivitydue to impurities.

a) To determine the resistivity of Be at 400 K, we can use the temperature coefficient of resistivity.

Given: Resistivity at 300 K (ρ₁) = 4 x 10^(-8) Ωm Temperature coefficient of resistivity (α) = 0.025 K^(-1)

We can use the formula: ρ₂ = ρ₁ * (1 + α * (T₂ - T₁))

Substituting the values: T₁ = 300 K T₂ = 400 K

ρ₂ = (4 x 10^(-8) Ωm) * (1 + 0.025 K^(-1) * (400 K - 300 K))

Simplifying the equation: ρ₂ = (4 x 10^(-8) Ωm) * (1 + 0.025 K^(-1) * 100 K)

ρ₂ = (4 x 10^(-8) Ωm) * (1 + 2.5)

ρ₂ = (4 x 10^(-8) Ωm) * 3.5

ρ₂ = 14 x 10^(-8) Ωm

Therefore, the resistivity of Be at 400 K is 14 x 10^(-8) Ωm.

b) To determine the contribution to resistivity due to impurities in a beryllium alloy containing 5 at% of an alloying element, we need to find the difference between the resistivity of the alloy and the resistivity of pure beryllium at 300 K.

Given: Resistivity of the alloy at 300 K (ρ_alloy) = 50 x 10^(-8) Ωm Resistivity of pure beryllium at 300 K (ρ_be) = 4 x 10^(-8) Ωm

Contribution to resistivity due to impurities = ρ_alloy - ρ_be

Contribution to resistivity due to impurities = (50 x 10^(-8) Ωm) - (4 x 10^(-8) Ωm)

Contribution to resistivity due to impurities = 46 x 10^(-8) Ωm

Therefore, the contribution to resistivity due to impurities in the beryllium alloy is 46 x 10^(-8) Ωm.