Examkin

Question

Difficulty: HardScientific Notation and Unit Conversions

During a nuclear physics experiment, a detector measures a neutron flux of 1.5×1012 neutrons per square millimeter per minute (neutrons/(mm2⋅min))1.5 \times 10^{12}\text{ neutrons per square millimeter per minute } (\text{neutrons}/(\text{mm}^2\cdot\text{min})). What is this neutron flux expressed in units of neutrons per square meter per second (\text{neutrons}/(\text{m}^2\cdot\text{s}))?

  1. A
    2.5×10132.5 \times 10^{13}
  2. B
    2.5×10102.5 \times 10^{10}
  3. 2.5×10162.5 \times 10^{16}Answer
  4. D
    9.0×10199.0 \times 10^{19}

Answer

The correct neutron flux is 2.5×1016 neutrons/(m2⋅s)2.5 \times 10^{16}\text{ neutrons}/(\text{m}^2\cdot\text{s}).
To convert the rate, we substitute the equivalent values for the units in the denominator: 1 mm2=10−6 m21\text{ mm}^2 = 10^{-6}\text{ m}^2 and 1 min=60 s1\text{ min} = 60\text{ s}. Subsituting these gives 1.5×1012 neutrons10−6 m2×60 s=1.5×10126.0×10−5 neutrons/(m2⋅s)=0.25×1017 neutrons/(m2⋅s)\frac{1.5 \times 10^{12}\text{ neutrons}}{10^{-6}\text{ m}^2 \times 60\text{ s}} = \frac{1.5 \times 10^{12}}{6.0 \times 10^{-5}}\text{ neutrons}/(\text{m}^2\cdot\text{s}) = 0.25 \times 10^{17}\text{ neutrons}/(\text{m}^2\cdot\text{s}), which simplifies to 2.5×1016 neutrons/(m2⋅s)2.5 \times 10^{16}\text{ neutrons}/(\text{m}^2\cdot\text{s}).

Step-by-Step Solution

1
Write the given rate expression as a ratio of quantities.
Rate = 1.5×1012 neutrons1 mm2⋅1 min\frac{1.5 \times 10^{12}\text{ neutrons}}{1\text{ mm}^2 \cdot 1\text{ min}}
This establishes the starting units that need to be converted in the denominator.
2
Convert the unit of area from square millimeters (mm2\text{mm}^2) to square meters (m2\text{m}^2).
1 mm2=(10−3 m)2=10−6 m21\text{ mm}^2 = (10^{-3}\text{ m})^2 = 10^{-6}\text{ m}^2
Since the unit is squared, the linear prefix conversion factor (10−310^{-3}) must also be squared.
3
Convert the unit of time from minutes to seconds.
1 min=60 s1\text{ min} = 60\text{ s}
This matches the target time unit in the denominator.
4
Substitute the converted unit values into the denominator of the rate expression.
Rate = 1.5×1012 neutrons(10−6 m2)⋅(60 s)\frac{1.5 \times 10^{12}\text{ neutrons}}{(10^{-6}\text{ m}^2) \cdot (60\text{ s})}
This replaces the original units with their equivalent values in the target units.
5
Simplify the expression to find the final value in scientific notation.
Rate = 1.5×10126.0×10−5 neutrons/(m2⋅s)=0.25×1017=2.5×1016 neutrons/(m2⋅s)\frac{1.5 \times 10^{12}}{6.0 \times 10^{-5}}\text{ neutrons}/(\text{m}^2\cdot\text{s}) = 0.25 \times 10^{17} = 2.5 \times 10^{16}\text{ neutrons}/(\text{m}^2\cdot\text{s})
Dividing the coefficient 1.51.5 by 6.06.0 yields 0.250.25, and subtracting the exponent in the denominator (−5-5) from the numerator (1212) yields 1717. Shifting the decimal point to standard scientific notation results in 2.5×10162.5 \times 10^{16}.

Key Concept

Scientific Notation and Unit Conversions
Rate this question

Topics

Scientific Notation and Unit Conversions

More questions from this topic