A circuit employs a silicon solar cell to detect flashes of light lasting 0.25 s. The smallest current the circuit can detect reliably is 0.32 μA .
Assuming that all photons reaching the solar cell give their energy to a charge carrier, what is the minimum power of a flash of light of wavelength 570 nm that can be detected?
Attempt at solution:
Using Equation 1: f=c/wavelength
=(3.00*10^8m/s)/(570*10^-9m)=5.26HZ
And Equation 2: E=hf
Planck's Constant= h=6.63*10^-34J*s= 4.14*10^-15eV*s
E=(6.63*10^-34J*s)(5.26HZ)=3.49*10^-19
3.49*10^-19/0.25s=1.39*10^-18 W
I'm not sure where to factor in the
0.32 μA, but I know that the unit is suppose to be in W.
To determine the minimum power of a flash of light that can be detected, we need to consider the smallest current the circuit can detect reliably.
The smallest current the circuit can detect reliably is given as 0.32 μA (microamperes). This current represents the flow of charge carriers (electrons) produced by the solar cell upon absorbing photons.
To relate the current to power, we can use the equation:
Power (P) = Current (I) * Voltage (V)
Since we are only given the current (I), we need to find the voltage (V) across the circuit. We can use Ohm's Law to do this:
Voltage (V) = Current (I) * Resistance (R)
However, the resistance (R) of the circuit is not given in the problem statement. Without the resistance value, we cannot directly calculate the voltage across the circuit. Therefore, we need to make an assumption.
Assumption: Let's assume that the resistance of the circuit is 1 Ohm. This assumption allows us to use the given current value (0.32 μA) to calculate the voltage across the circuit.
Using Ohm's Law, we find:
Voltage (V) = 0.32 μA * 1 Ohm = 0.32 μV (microvolts)
Now, to convert this voltage to power, we use the equation:
Power (P) = Voltage (V) * Current (I)
Power (P) = (0.32 μV) * (0.32 μA) = 0.1024 μW (microwatts)
Therefore, the minimum power of a flash of light that can be detected by the circuit is approximately 0.1024 μW.
To factor in the 0.32 μA (microamperes), you need to calculate the power (P) based on the current (I).
Given:
Current (I) = 0.32 μA = 0.32 x 10^(-6) A
Power (P) can be calculated using the equation: P = IV, where I is the current and V is the voltage.
Since the solar cell only detects flashes of light, we can assume that the voltage across the solar cell is constant.
Therefore, to calculate the minimum power of a flash of light that can be detected, you need to multiply the current (I) by the constant voltage.
Let's assume V = 1 V (for the sake of simplicity, as the exact value of voltage is not given in the problem).
P = 0.32 x 10^(-6) A x 1 V
P = 0.32 x 10^(-6) W
So, the minimum power of a flash of light that can be detected is 0.32 x 10^(-6) W.