The frequency of the current observed at a Josephson junction is 976 MHz.The voltage applied to the junction to produce this current was
A)202 µV
B)1.43 V
C)1.43 µV
D)2.02 µV
E)1.28 µV

Question

Quizplus · Accepted Answer

The frequency of the current observed at a Josephson junction is related to the voltage applied across it by the Josephson frequency-voltage relation, which is given by $f = \frac{2eV}{h}$, where $f$ is the frequency, $V$ is the voltage, $e$ is the elementary charge ($1.6 \times 10^{-19}$ coulombs), and $h$ is Planck's constant ($6.626 \times 10^{-34}$ Js). Rearranging the formula to solve for $V$, we get $V = \frac{fh}{2e}$. Substituting the given frequency $f = 976 \times 10^6$ Hz, $h = 6.626 \times 10^{-34}$ Js, and $e = 1.6 \times 10^{-19}$ C, we find $V = \frac{976 \times 10^6 \times 6.626 \times 10^{-34}}{2 \times 1.6 \times 10^{-19}}$ volts, which calculates to approximately 2.02 µV.

The Frequency of the Current Observed at a Josephson Junction