Question

The energy released by the fission of one uranium atom is $200\mathrm{MeV}$. The number of fission Per second required to Produce $3.2\mathrm{w}$ of Power is (A) ${10}^{10}$ (B) ${10}^7$ (C) ${10}^{12}$ (D) ${10}^{11}$

Short answer

The number of fission per second required to produce 3.2 W of power is approximately ${10}^{12}$ fissions/s (Option C).

Step-by-step solution

Convert the energy released per fission to Joules

We are given that the energy released by the fission of one uranium atom is 200 MeV. To convert MeV to Joules, we use the conversion factor 1 MeV = $1.602\times {10}^{-13}$ Joules. So, the energy released per fission is: 200 MeV * $(1.602\times {10}^{-13}){\displaystyle \frac{J}{MeV}}$ = $3.204\times {10}^{-11}$ Joules.

Convert the power output to Joules per second

The given power output is 3.2 W. Since 1 Watt is equal to 1 Joule per second, the power output is already in the required unit. Power output = 3.2 J/s.

Use the formula Power = Energy / Time to find the number of fissions required per second

To find the number of fissions per second required, we will use the formula Power = Energy / Time. We have the power output and the energy released per fission, so we can plug these values into the formula and solve for the number of fissions: 3.2 J/s = $(3.204\times {10}^{-11}\mathrm{J}/\mathrm{fission})\times (\mathrm{number}\mathrm{of}\mathrm{fissions}/\mathrm{s})$ Now, we'll solve for the number of fissions per second: $\mathrm{number}\mathrm{of}\mathrm{fissions}/\mathrm{s}$ = ${\displaystyle \frac{3.2\mathrm{J}/\mathrm{s}}{3.204\times {10}^{-11}\mathrm{J}/\mathrm{fission}}}$ = $9.98753\times {10}^{10}$ fissions/s.

Choose the closest answer from the given options

Since the number of fissions per second required is approximately $9.98753\times {10}^{10}$ fissions/s, the closest option among the given choices is: (C) ${10}^{12}$ fissions/s.

Key concepts

Energy Conversion

Energy conversion is about changing energy from one form to another. In the case of nuclear fission, it involves converting nuclear energy into a form we can measure and use, like Joules. During fission, an atom like uranium is split into smaller parts, releasing a significant amount of energy. This energy is initially measured in mega-electronvolts (MeV), which is a unit used to denote energy on an atomic scale. The challenge is, our day-to-day energy measurements are typically in Joules, not MeV. So, we perform a conversion: 1 MeV is equivalent to approximately $1.602\times {10}^{-13}$ Joules. By using this conversion factor, we can calculate the energy released by the fission of one uranium atom in more familiar terms, helping to understand the power output more intuitively.

Power Calculation

Power tells us how much energy is being used or produced per unit time. It is usually measured in Watts, where 1 Watt is equal to 1 Joule per second. When calculating power in a nuclear reaction, we need to consider how much energy each fission event releases and how many of these events occur per second. In our exercise, we are asked to find out how many fission events happen every second to yield a power output of 3.2 Watts (or 3.2 Joules per second). To do that, once we know the energy per fission in Joules, we can use the formula:

• Power = Energy/Time

With this, we can solve for the time in terms of the number of fission events calculated per second, giving us insight into the energy dynamics of nuclear power.

Joule

Joule is the standard unit of energy in the International System of Units (SI). Named after physicist James Prescott Joule, this unit allows us to quantify energy in various contexts like thermal, mechanical, or nuclear energy. The essential role of the Joule is to provide a consistent measure you can use to compare different forms of energy. In nuclear fission, the energy released by atomic reactions is often translated from MeV to Joules so it can be conveniently understood alongside all other forms of energy in common calculations. This conversion aids not just in theoretical physics, but in practical applications like providing energy rates in familiar units like Watts, so we understand the power these reactions can produce.

MeV

MeV, or mega-electronvolt, is a unit of energy used primarily in atomic and nuclear physics. It represents one million electronvolts. The MeV is particularly useful when discussing nuclear processes because it corresponds well to the energy scales naturally occurring in these reactions. Converting from MeV to more familiar units like Joules is crucial since many practical applications and power requirements in daily life are described in Joules or related units like Watts. In the context of nuclear fission, noting the energy in MeV and then converting to Joules enables engineers and scientists to plan, control, and optimize energy production from nuclear reactions to match needed power outputs in settings including power plants and research facilities.