The charge (in coulombs) on an object increases according to (q)=(2 coul)e^[0.05/sec)(t)], where t is in seconds. How many seconds is the object deficient 1.56x10^20 electrons (from neutral)?
0, 12.56, 21.94, 41.21, 28.0, 50.5, 65.90, 105.00, 874.00, 1000.00
To find the number of seconds the object is deficient by 1.56x10^20 electrons, we need to determine the charge on 1.56x10^20 electrons and then solve for t in the given equation.
First, we need to know the charge on a single electron. The elementary charge is equal to 1.602 x 10^-19 coulombs. Therefore, the charge on 1.56x10^20 electrons would be:
Charge = (1.56x10^20) x (1.602 x 10^-19) C
Charge = 2.49912 C
Now, we can set up the equation and solve for t:
2.49912 = (2 e^(0.05t))
Let's isolate the exponential term first by dividing both sides by 2:
1.24956 = e^(0.05t)
Next, we take the natural logarithm (ln) of both sides to remove the exponential:
ln(1.24956) = ln(e^(0.05t))
Using the property of logarithms, ln(e^(0.05t)) simplifies to:
0.05t = ln(1.24956)
Now, divide both sides by 0.05 to isolate t:
t = (ln(1.24956)) / 0.05
Calculating this using a scientific calculator or computing software, we get t ≈ 12.56 seconds.
Therefore, the object is deficient by 1.56x10^20 electrons after approximately 12.56 seconds.
So, the correct answer from the given options is 12.56 seconds.