The energy required for self-diffusion is 1.6eV. At 100 degrees C, 1% of the atoms have the energy required for self-diffusion. How many of the atoms would have the energy required for self-diffusion at 120 degrees C?

12.6% of the atoms

Explanation:

We know that :

In (D2/D1) = [Qd/R] [(1/T1) – (1/T2)]

where Qd is activation energy for diffusion

T is temperature

Given that

Qd = activation energy for diffusion = 1.6 eV

R = 8.62 x 10-5 eV/K

D1 = 1 %

D2 = ?

T1 = 100°C = 100 + 273 K = 373 K

T2 = 120°C = 120 + 273 K = 393 K

Then,

Calculation:

In (D2/D1) = [Qd/R] [(1/T1)] – (1/T2)]

In (D2/1) = [1.6 eV /8.62 x 10⁻⁵ eV/K] [(1/373)] – (1/393)]

On solving,

D2 = 12.6 %

Therefore,

12.6% of the atoms have the energy required for self-diffusion at 120°C.

chemistry sucks a**

explanation: