For the temperatures of 77 degree K and 42 degree K calculat

For the temperatures of 77 degree K and 4.2 degree K, calculate the equivalent thermal energy kT. Based on this, calculate the capacitance needed to have the same capacitive energy (E_c) for a single electron (charge on the capacitor is e). Assume as we did in class that the relative dielectric constant is 10 and the distance between capacitor plates (d) is about 1 nm. Calculate the areas of the capacitors needed to observe the effects of a single electron at these two temperatures. These temperatures are significant because they are the temperatures of liquid nitrogen and liquid helium respectively. These capacitances are easier to fabricate, but it is not economical to operate electronic devices at low temperatures. Consider the current -voltage coulomb staircase data below. Answer the following. What capacitance does this correspond to? What is the charging energy associated with this system? What is the approximate effective resistance of the most resistive tunnel junction in this system? From #4 you calculated the thermal energy associated with 77 degree K. Compare this with the answer to part a. How does this help you understand the data given?

Solution

Answer 4

Here, k is Boltzmann constant = 1.38064852 × 10²³ JK¹

        And T is temperature

E= 1/2Q²/C

here, C is capacitance and Q is charge (which according to question is

e = 1.60217662 × 10^-19 coulombs )

so, C= 2E/Q²

1. for T= 4.2 ºK: C= 2 x 5.798723784 x 10²³ / (1.60217662 × 10^-19)² = 4.517952263513133 x 10¹⁵ Farad

2. for T= 77 ºK: C= 2 x 106.30993604 x 10²³ / (1.60217662 × 10^-19)² = 82.82912483107411 x 10¹⁵ Farad

so, Area of capacitor (A) = C x d/ k_o

1. for T= 4.2 ºK: A = 0.5102 x 10¹⁴ sq. m

2. for T= 77 ºK: A = 9.354 x 10¹⁴ sq. m