The frequency of the current observed at a Josephson junction is 345 MHz.The voltage applied to the junction to produce this current was
A)1.43 µV
B)1.43 V
C)0.715 µV
D)0.715 mV
E)1.28 µV

Question

Quizplus · Accepted Answer

The frequency of the current in a Josephson junction is related to the voltage applied by the following equation: f = 2eV/h, where e is the charge of an electron, V is the voltage, and h is Planck's constant.

Solving for V, we get: V = fh/2e = (345 MHz)(6.626 x 10^-34 J⋅s)/(2 x 1.602 x 10^-19 C) ≈ 0.715 µV.

Therefore, the voltage applied to the junction to produce this current was 0.715 µV, which corresponds to choice C.

The Frequency of the Current Observed at a Josephson Junction