The laser system tested for inertial confinement can produce a \(100 - kJ\) pulse only \(1.00{\rm{ }}ns\) in duration.

(a) What is the power output of the laser system during the brief pulse?

(b) How many photons are in the pulse, given their wavelength is \(1.06{\rm{ }}\mu m\)?

(c) What is the total momentum of all these photons?

(d) How does the total photon momentum compare with that of a single \(1.00{\rm{ }}MeV\) deuterium nucleus?

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

• Pulse of laser system is:\(100 - kJ\).
• Duration for production of pulse:\(1.00{\rm{ }}ns\).
• Wavelength of photon is:\(1.06{\rm{ }}\mu m\).
• Energy of deuterium nucleus is: \(1.00{\rm{ }}MeV\).

The ratio of energy and time will give the power –

\(P = \frac{E}{t}\)

Putting in the values from the –

\(P = \frac{{100 \times {{10}^3}J}}{{1.00 \times {{10}^{ - 9}}s}}\)

And the result is obtained as –

\(P = 1.00 \times {10^{14}}{\rm{ }}W\)

Therefore, the value for power is obtained as \(P = 1.00 \times {10^{14}}{\rm{ }}W\).

From the Planck-Einstein relation from the quantization of the photon,the energy of photon can be deduced –

\({E_{photon{\rm{ }}}} = \frac{{hc}}{\lambda }\)

And to get the number of photons simply divide full pulse energy by the energy of each photon –

\(N = \frac{E}{{{E_{photon{\rm{ }}}}}}\)

Rearranging the equation –

\(\begin{array}{c}N = \frac{E}{{\frac{{hc}}{\lambda }}}\\ = \frac{{E\lambda }}{{hc}}\end{array}\)

Now putting in all of the numbers, it is obtained –

\(\begin{array}{c}N = \frac{{\left( {100 \times {{10}^3}\;J} \right)\left( {1.06 \times {{10}^{ - 6}}\;m} \right)}}{{\left( {6.63 \times {{10}^{ - 34}}\;J \cdot s} \right)\left( {3.00 \times {{10}^8}\;m/s} \right)}}\\N = 5.3 \times {10^{23}}\end{array}\)

Therefore, the value for number of photons is obtained as \(N = 5.3 \times {10^{23}}\).

The total momentum of these photons can be calculated by relation from the De-Broglie –

\(p = \frac{h}{\lambda }\)

But since we it is asked to calculate the total momentum of all these photons the relation with the energy can be used theenergy of the whole pulse is taken –

\(p = \frac{E}{c}\)

Putting in the values from the –

\(\begin{array}{c}p = \frac{{100 \times {{10}^3}\;J}}{{3.00 \times {{10}^8}\;m/s}}\\p = 3.33 \times {10^{ - 4}}\;kg \cdot \frac{m}{s}\end{array}\)

Therefore, the value for momentum of photons is obtained as \(p = 3.33 \times {10^{ - 4}}\;kg \cdot \frac{m}{s}\).

The definition of momentum suggests –

\(p = mv\)

The definition of kinetic energy suggests –

\({E_k} = \frac{1}{2}m{v^2}\)

The calculation is on atom (molecule) level, so take into account the molar mass, or simply with Avogadro constant –

\(m = \frac{{NM}}{{{N_A}}}\)

Since it is only one deuterium it can be written –

\(m = \frac{M}{{{N_A}}}\)

So, the momentum relation becomes –

\(p = \frac{M}{{{N_A}}}v\)

Before calculating the momentum, the speed of deuteron needs to be determined, which can be done from kinetic energy relation –

\({E_k} = \frac{1}{2}m{v^2}\)

And taking a square root for the speed gives –

\(v = \sqrt {\frac{{2 \cdot {E_k}}}{m}} \)

Putting in the numbers gives –

\(\begin{array}{c}v = \sqrt {\left[ {\frac{{2\left( {1.60 \times {{10}^{ - 13}}\;J} \right)}}{{\left( {2.014 \times {{10}^{ - 3}}\;kg} \right)/\left( {6.022 \times {{10}^{23}}} \right)}}} \right]} \\v = 9.78 \times {10^6}\;m/s\end{array}\)

And now putting this into momentum relation –

\(\begin{array}{c}p = \frac{M}{{{N_A}}}v\\ = \left( {\frac{{2.014 \times {{10}^{ - 3}}\;kg}}{{6.022 \times {{10}^{23}}}}} \right)\left( {9.78 \times {{10}^6}\;m/s} \right)\\p = 3.27 \times {10^{ - 20}}\;kg\frac{m}{s}\end{array}\)

Calculate the ratio of the momentum of the single-photon to the momentum of the deuterium –

\(\begin{array}{c}\frac{{{\rm{ }}{p_{photon}}{\rm{ }}}}{{{p_D}}} = \frac{{3.333 \times {{10}^{ - 4}}\;kg \cdot m/s}}{{3.27 \times {{10}^{ - 20}}\;kg \cdot m/s}}\\ = 1.019 \times {10^{16}}\end{array}\)

Therefore, the ratio of momentum of photon and single deuterium nuclei is\(\frac{{{\rm{ }}{p_{photon}}{\rm{ }}}}{{{p_D}}} = 1.019 \times {10^{16}}\).