Question

If a processor clock is rated as 2500 million cycles per second, then its clock period is:

• A. \( 2.50 \times 10^{-10} \) sec
• B. \( 4.00 \times 10^{-10} \) sec
• C. \( 1.00 \times 10^{-10} \) sec
• D. None of the above

Hint:

Remember the fundamental relationship: Period = 1 / Frequency. For processor speeds, "million cycles per second" is MHz (\(10^6\)) and "billion cycles per second" is GHz (\(10^9\)). 2500 million is 2.5 billion, or 2.5 GHz.

Answer 1

The Correct Option is B

The clock period (T) is the time for one clock cycle, and it is the reciprocal of the clock frequency (f).

Step 1: Express the frequency in Hertz (cycles per second). The given frequency is 2500 million cycles per second.
\( f = 2500 \times 10^6 \) Hz = \( 2.5 \times 10^9 \) Hz (or 2.5 GHz).

Step 2: Calculate the period.
The period is the inverse of the frequency: \( T = \frac{1}{f} \). \[ T = \frac{1}{2.5 \times 10^9} \text{ seconds} \] \[ T = \frac{1}{2.5} \times 10^{-9} \text{ seconds} \]
To simplify \( \frac{1}{2.5} \), we can write it as \( \frac{10}{25} = \frac{2}{5} = 0.4 \).
So, \( T = 0.4 \times 10^{-9} \) seconds. To express this in standard scientific notation (with one non-zero digit before the decimal), we write:
\[ T = 4.0 \times 10^{-1} \times 10^{-9} = 4.0 \times 10^{-10} \text{ seconds} \] This matches option (B).