Q1: The diffusion co-efficient of Ni in Cu at 1000 K is 1.93 x 10^-16 m² . S^-1 and it is 1.94 x 10^-14 m² S^-1 at 1200 K. The activation energy (in k. J.mole^-1 ) for the diffusion of Ni in Cu is

ANS:C - 230

To find the activation energy for the diffusion of Ni in Cu, we can use the Arrhenius equation: D=D0​exp(−RTQ​) Where:

• D is the diffusion coefficient.
• D0​ is the pre-exponential factor.
• Q is the activation energy.
• R is the gas constant.
• T is the temperature in Kelvin.

Given that the diffusion coefficient (D) changes from 1.93×10^−16m2⋅s^−1 at 1000K to 1.94×10−14 m2⋅s−1 at 1200 , we can set up two equations using the Arrhenius equation: D1​=D0​exp(−RT1​Q​) D2​=D0​exp(−RT2​Q​) Where D1​ and D2​ are the diffusion coefficients at temperatures T1​ and T2​ respectively. Taking the ratio of these two equations: D1​D2​​=D0​exp(−RT1​Q​)D0​exp(−RT2​Q​)​ D1​D2​​=exp(−RT1​Q​)exp(−RT2​Q​)​ D1​D2​​=exp(RQ​(T1​1​−T2​1​)) Now, we can rearrange this equation to solve for Q, the activation energy: ln(D1​D2​​)=RQ​(T1​1​−T2​1​) Q=R(T1​1​−T2​1​ln(D1​D2​​)​) Given that R=8.314J⋅K−1⋅mol−1, we can substitute the values: T1​=1000K T2​=1200K D1​=1.93×10−16m2⋅s−1 D2​=1.94×10−14m2⋅s−1 Let's calculate Q using these values. Using the given values: T1​=1000K T2​=1200K D1​=1.93×10−16m2⋅s−1 D2​=1.94×10−14m2⋅s−1 And R=8.314J⋅K−1⋅mol−1, we can calculate Q: Q=8.314×10−3×10001​−12001​ln(1.93×10−161.94×10−14​)​ Let's calculate Q: Q=8.314×10−3×10001​−12001​ln(1.93×10−161.94×10−14​)​ Q=8.314×10−3×1000×12001200−1000​ln(1.93×10−161.94×10−14​)​ Q=8.314×10−3×1200000200​ln(1.93×10−161.94×10−14​)​ Q=8.314×10−3×200ln(1.93×10−161.94×10−14​)×1200000​ Q≈8.314×10−3×200ln(100)×1200000​ Q≈8.314×10−3×2004.605×1200000​ Q≈8.314×10−3×276.3 Q≈2.298kJ/mol So, the activation energy for the diffusion of Ni in Cu is approximately 2.298 kJ/mol2.298kJ/mol.