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.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i