The Science Thinkers: Atomic Physics

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Nuclear Battery

Atomic Physics, Battery, Nuclear Battery

Batteries are necessities of modern life, powering everything from cell phones to vehicles. Most batteries operate via chemical reactions that convert stored chemical energy into electrical energy. These electrochemical reactions are triggered when a load (e.g. a light bulb) is connected to a battery, causing electrons to flow.

Though traditional batteries are ubiquitous and can be very cheap, future batteries may harness the power of radioactive isotopes for electricity generation. Batteries that use the decay products of radioactive isotopes are known as atomic batteries or radioisotope generators.

The first atomic battery was developed in 1913 by Henry Moseley. His battery consisted of a spherical glass globe with a silver lining the interior. Within the glass, the sphere was an emitter of a radioactive isotope of radium. The emitted charged particles deposited on the silver, causing a build-up of charge. This effectively created a radioactively powered capacitor, from which electric current could be extracted.

The overwhelming advantage of atomic batteries is their long life capabilities with minimal external maintenance. Therefore, radioisotope generators are ideal for space missions lasting several years or for power generation in remote locations. Furthermore, atomic batteries are lightweight, independent of sunlight unlike solar cells and unaffected by radiation belts in space (e.g. the Van Allen belts). Nonetheless, atomic batteries have only proliferated slowly primarily due to their prohibitive cost compared to traditional batteries and public health concerns regarding radioactivity.

Just as electrochemical batteries can be fueled by reactions between various chemicals, atomic batteries can be fueled by the emissions of many radioactive isotopes. However, the electric conversion principles employed distinguish atomic batteries into two categories: thermal and non-thermal. Power output from a thermal atomic battery is dependent on temperature whereas a non-thermal battery is independent of temperature.

NASA and the Department of Energy have extensively studied radioisotope power systems to produce heat and electricity for space missions lasting over a decade and extending into regions of the solar system where sunlight is too faint to permit solar energy conversion as a viable power source. As of 2005, the United States had launched 44 radioisotope thermoelectric generators (RTGs) on 25 different missions, including the Cassini mission to Saturn and the Galileo mission to Jupiter.

The radioisotope of primary interest to NASA is plutonium-238 (Pu238), a radioisotope not used for nuclear weapons. Plutonium decay products not only serve to generate electricity to power the spacecraft, but the heat generated by radioactive decay fuel radioisotope heater units which warm the instruments aboard the spacecraft. Keeping electronics and instruments warm is essential in the frigid climate of space since most electronic equipment has a relatively narrow operating temperature range.

NASA and the DOE continue to fund research on RTGs with goals to increase power conversion efficiency above 35% while maintaining the reliability of electricity generation in space. More than ten government contracts have been awarded to private companies relating to various aspects of RTGs in an effort to produce more cost-effective long-term science missions for NASA.

A non-thermal atomic battery that generates electricity from electron/beta-particle emission is known as a betavoltaic. A viable radioisotope for use in betavoltaics is an isotope of hydrogen with two additional protons, known as tritium. Since tritium loses half its radioactive energy in 12.3 years (its half-life) betavoltaics made with tritium can last well over a decade.

Furthermore, beta particles emitted from tritium do not penetrate human skin, alleviating potential public health concerns. These two characteristics of tritium make it well-suited for medical devices. For example, pacemaker recipients often outlive the pacemaker battery, requiring risky, invasive replacement surgery. Pacemakers using a tritium power source could have battery lifetimes on the order of decades, obviating the need for replacement surgery.

Another advantage of tritium is that can be easily obtained from Canadian nuclear reactors that generate heavy water as a by-product. Though betavoltaics is still more expensive than other traditional batteries, they could provide a future power source for electronics requiring minimal power in poorly accessible locations.

Also Read:- What Is Radioisotope Power System?

What Would Happen If You Detonate A Nuclear Bomb In Space?

Atomic Physics, Nuclear physics, particle Physics, Physics, Science, Space

Detonating a nuclear bomb seems like a risky business in general, but in the early 1960s, the US and the Soviet Union were busy trying to figure out what would happen if you set one off in space? The answer turned out to be something they didn't really expect. A nuclear blast could cause a high-altitude electromagnetic pulse or EMP, a powerful man-made burst of electromagnetic energy that could basically wipe out communications here on Earth.

It was a whole new way to use the bomb and they kind of discovered it by accident. It all started in 1958 when the US launched its first satellite Explorer 1. NASA scientist James Van Allen equipped it with a Geiger counter, because he wanted to measure radiation at different altitudes, a project he had already been working on using balloons.

The readings that came back were strange. Radiation levels seemed to increase with altitude, then suddenly dropped to 0, then increase and then suddenly dropped to 0 again. More testing showed that readings that looked like 0 were actually because the radiation levels were so high that the detector couldn't handle it. But that spring Van Allen realized he made a new discovery: that there were at least two belts around the planet between 1,000 and 60,000 kilometres up with extra high concentrations of charged particles like electrons and protons.

Today we call these belts the Van Allen belts and we know that they are mostly made of particles from solar wind and cosmic rays, held in place by Earth's magnetic field. We also know that depending on solar activity there can be more than two of them.

In May of that year, Van Allen presented his discovery at a press conference at the National Academy in Washington DC. Later that day, the US military asked for his help detonating nuclear weapons in the Van Allen belts. US military officials suspected the Soviets were doing high-altitude nuclear tests. And at the time nobody really knew how a high-altitude explosion would differ from one here on Earth. Newly discovered Van Allen Belts added a whole new element because nuclear blasts release lots of charged particles and here with these two huge bands of more charged particles.

The US was worried that interference from the Van Allen belts might hide incoming missiles or that they could somehow be used to steer a blast. So they decided to learn more about how atomic bombs behaved at high altitudes by detonating a bunch of them. During those tests they ended up measuring electric signals so high they thought it was a fluke caused by other flaws in the instruments they were using. But they had to wait a while to figure out what was really happening, because later that year the USSR called for a ban on high-altitude nuclear testing and the US agreed.

Then in 1961, the USSR started testing their own nukes at high altitudes anyway and the US quickly continued their own, including a test known as Starfish Prime. Starfish Prime was humanity's first hydrogen bomb detonated at a high altitude. It detonated 400 kilometres above a point near its launch from Johnston Island in the Pacific Ocean. It was also the biggest bomb ever set off in space, 100 times more powerful than the Hiroshima bomb and with a blast equal to about 1.4 million tons of TNT.

But with so little air around it, it didn't make a fireball. Instead, the charged particles zooming away from Starfish Prime caused a huge Aurora that could be seen for thousands of kilometres around Johnston Island. And then, burglar alarms started going off in Hawaii more than a thousand kilometres away. 27 rockets followed Starfish Prime to gather data and even they weren't equipped to measure what happened. What the US learned was that the oddly high measurement from earlier tests weren't glitches.

High-altitude nuclear explosions are just very different. In the near vacuum up there, the energy from nuclear blast sends out lots of free electrons. Those electrons create a brief but extremely powerful electromagnetic pulse: an EMP. Starfish Prime's EMP was so strong, it affected the flow of electricity on Earth thousands of kilometres from Johnston Island, causing blackouts and electrical malfunctions in Hawaii and disabling at least six satellites. But that was in a relatively isolated area.

Today an EMP could be used to disable an entire country. The US commission to study EMPs in 2008 estimated that an EMP attack could kill 90% of the US population within 12 months since so much of the way we live depends on satellites and the electrical grid. The US military ran a few more tests after Starfish Prime, but they kept things a lot smaller and the data from those tests is still classified.

Only a year later the USSR proposed another moratorium on high-altitude nukes: the Limited Test Ban Treaty of 1963, and we haven't set off any nukes in space since then. So in the end humanity was probably right to be worried about the consequences of detonating nuclear bombs in space, but mainly because they accidentally stumbled upon a way to make the aftermath of the bomb even worse.

Rutherford's Alpha particle scattering Experiment

Atomic Physics No comments

Rutherford's Alpha( $\alpha-$ ) Particle Scattering

The theory of rutherford's $\alpha-$ particle scattering is based on the

Following Assumption:-

The entire charge and almost entire mass of the atom is concentrated in a single core of the atom is called nucleus and is surrounded by Negatively(-) charged electron clouds.
The $\alpha-$ particle and gold nuclei are so small that they may be treated as a point mass and point charge.
The nucleus is considered so heavy that its motion during the scattering process may be neglected.
The scattering is due to the coulombs force of repulsion between the $\alpha-$ particle and the Gold Nucleus.
Each $\alpha-$ particle suffers a single deflection.
The $\alpha-$ particles do not penetrate the nuclear region. So strong nuclear forces are not involved in their interaction.

Derivation Of Rutherford's Scattering Formula

Let an $\alpha-$ particle is moving along PO apparatus and the relatively heavy nucleus stationary at it. Since both are Positively(+) charged so there is a force of repulsion between them. As $\alpha-$ particle gets closer to the nucleus the repulsive force highly increases and the $\alpha-$ particle follows a hyperbolic path. The asymptotes and po and p'o give the initial and final reflection of $\alpha-$ particle.
Let the perpendicular drawn from the nucleus to PO is NM and is the shortest distance between the nucleus and initial direction of 𝞪- particle which is known as impact parameters(b).

Let

Z be the atomic no. of the element That scattered $\alpha-$ particle.
Ze be the charge of the nucleus.
θ, be the scattering angle or angle of deviation of $\alpha-$ particle.
As the $\alpha-$ particle moves its momentum changes from $\vec{P_{1}}$ to $\vec{P_{2}}$ i.e $\vec{\Delta P}= \vec{P_{2}}-\vec{P_{1}}=\int \vec{F}dt------(1)$

using sine law in $\Delta zxy$

$\frac{yz}{sin\Theta }=\frac{xz}{sin\frac{\pi -\theta }{2}}$
$\Rightarrow \frac{\Delta P}{sin\theta }=\frac{\vec{P_{2}}}{sin(\frac{\pi }{2}-\frac{\theta }{2})}$

$\Rightarrow \frac{\Delta P}{sin \theta }=\frac{mv}{cos(\frac{\theta}{2})}$

$\Rightarrow \Delta P=\frac{mvsin\theta}{cos(\frac{\theta}{2})}$

$\Rightarrow \Delta P=\frac{2mv . Sin(\frac{\theta}{2}) cos(\frac{\theta}{2})}{cos(\frac{\theta}{2})}$

${\color{Red}\Rightarrow \Delta P= 2.mv.sin(\frac{\theta}{2})........(2) }$

Now From Equation (1)

$\Rightarrow \Delta P =\int_{-\infty }^{\infty}Fdt$

$\Delta P=\int_{-\infty }^{\infty }F.cos\phi.dt$

where t = $- \infty$ ,    $\phi= -(\frac{\pi -\theta }{2})$

t = $\infty$ ; $\phi = (\frac{\pi -\theta }{2})$
$\Rightarrow \Delta P = \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}} Fcos\phi dt$

$\Rightarrow \Delta P = \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}} Fcos\phi \frac{dt}{d\phi }.d\phi$

From Eq (2)

${\color{Red} \Rightarrow 2mvsin\frac{\theta }{2} = \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}} Fcos\phi \frac{dt}{d\phi }.d\phi--------------(3)}$

As the radius   $\vec{b}$ joining the $\alpha$ -- particle and the nucleus is along the direction of the force. So there is no torque acting on the $\alpha$ -- particle.

The angular momentum remain content

$\Rightarrow I\omega =L$

$\Rightarrow m r^{2}\omega =bmv$ [ L=R *P =Rmv=bmv ]

$\Rightarrow \frac{d\theta }{dt}=\frac{bv}{r^{2}}$

$\Rightarrow \frac{dt}{d\theta}=\frac{r^{2}}{bv}$

putting this   $\theta$ in EQ (3)

${\color{Red} 2mv sin\frac{\theta }{2}=\int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}}Fr^{2}cos\phi d\phi ---------(4)}$

But    $F=\frac{1}{r\pi \epsilon_{0} }\frac{Ze2e}{r^{2}}$

   $\Rightarrow F=\frac{1}{4\pi\epsilon _{0}}\frac{2Ze^{2}}{r^{2}}$

putting this EQ (4)

$\Rightarrow 2mbv^{2}sin(\frac{\theta }{2})= \int_{-(\frac{\pi -\theta }{2})}^{\frac{\pi -\theta }{2}}\frac{1}{4\pi \epsilon _{0}}\frac{2Ze^{2}}{r^{2}} r^{2}cos\phi d\phi$

$\Rightarrow mbv^{2}\sin \frac{\theta }{2}=\frac{Ze^{2}}{4\pi \epsilon _{0}}\int_{-(\frac{\pi -\theta }{2})}^{(\frac{\pi -\theta }{2})}\cos \phi d\phi$

$\Rightarrow mbv^{2}\sin (\frac{\theta }{2})=\frac{Ze^{2}}{4\pi \epsilon _{0}}\left [ \sin (\frac{\pi -\theta }{2})-\sin- (\frac{\pi -\theta }{2}) \right ]$

$\Rightarrow mbv^{2}\sin (\frac{\theta }{2})=\frac{Ze^{2}}{4\pi \epsilon _{0}}\left ( \cos \frac{\theta }{2} +\cos \frac{\theta }{2}\right )$

$\Rightarrow mbv^{2}\sin (\frac{\theta }{2})=\frac{Ze^{2}}{4\pi \epsilon _{0}}2\cos \frac{\theta }{2}$

$\Rightarrow \frac{mbv^{2}.4\pi \epsilon _{0}}{2Ze^{2}}=\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=\cot \frac{\theta }{2}$

$\Rightarrow \cot \frac{\theta }{2}=\frac{2\pi \epsilon _{0}mv^{2}b}{Ze^{2}}=\frac{2\pi \epsilon _{0}.2.\frac{1}{2}.mv^{2}b}{Ze^{2}}$

${\color{DarkRed} \Rightarrow \cot \frac{\theta }{2}=\frac{4\pi \epsilon _{0}kb}{Ze^{2}}--------------(5)}$    where k=K.E

This is the relation between scatting angle and impact parameter.

In the actual experiment, a large number of $\alpha-$ particle are incident various impact parameter all around the (Au) Gold nucleus.
According to EQ(5), the $\alpha-$ particle that approaches the nucleus with impact parameter "b" will be scattered at an angle " $\theta$ "

$\alpha-$ Particle approaching with the smallest value of "b" (i.e $\leq$ b) will be scattered through large angle (i.e $\geq \theta$ )
The Area of cross-section of radius it.e $\pi b^{2}$ is called scattering cross-section i.e   ${\color{Red} \sigma = \pi b^{2}}$
Consider a thin gold foil of thickness "t" and cross - section "A". Suppose it contains 'n' number of atoms per unit volume, then the volume of the foil is "At".

Then the number of target nuclei in the foil is "nAt". We assume that the foil is so thin that $\alpha-$ particle suffers single deflection i.e , one $\alpha-$ particle is scattered by one gold nucleus.

Since one nucleus has cross - section $\LARGE \sigma$ .
So area of cross - section of "nAt" nuclei = $\LARGE \sigma nAt$ .

Let 'N' be the total number of $\alpha-$ particle and " $\large N_{0}$ " be the Number of $\alpha-$ particle being scattered by an angle $\geq \theta$ Then the fraction of incident $\alpha-$ particles scattered by an angle $\geq \theta$ is

$\large f=\frac{N}{N_{0}}=\frac{target\: area\: exposed\: by\: scattere}{total\: target\: area}$

$\large \Rightarrow f=\frac{\sigma nAt}{A}$

$\large \Rightarrow f=\sigma nt$

$\large \Rightarrow f=\pi b^{2}nt$ as $\large \sigma =\pi b^{2}$

$\large \Rightarrow f=\pi nta^{2}cot^{2}\frac{\theta }{2}$
-------------------------------------------
as we know from Eq(5)
   $\large \left | \cot\frac{\theta }{2}=\frac{4\pi \epsilon _{0}kb} {Ze^{2}}\Rightarrow b=\frac{Ze^{2}cot\frac{\theta }{2}}{4\pi \epsilon _{0}k} \right |$

   $\large \left | \Rightarrow b=a\cot\frac{\theta }{2} \left ( a=\frac{Ze^{2}}{4\pi \epsilon _{0}k} \right ) \right |$
-------------------------------------------
But a particle detector measures the Number of df i.e,

$df=\frac{dN}{N_{0}}$

$\Rightarrow df=\pi nta^{2}.2cot\frac{\theta }{2}(-cosec^{2}\frac{\theta }{2}).\frac{1}{2}d\theta$

$\Rightarrow df=-\pi nta^{2} \frac{cot\frac{\theta }{2}}{sin^{2}\frac{\theta }{2}} d\theta$

${\color{Red} \Rightarrow \frac{dN}{N_{0}}=-\pi nta^{2} \frac{cot\frac{\theta }{2}}{sin^{2}\frac{\theta }{2}} d\theta }$

The scattered   $\alpha-$ particle strikes the screen placed at a distance "r" from the foil. The area ds is struck by these particles is the area of the ring of radius rsin $\theta$ and thickness " $rd\theta$ " is
$ds= 2\pi (rsin\theta )rd\theta =2\pi r^{2}sin\theta d\theta$

$ds=2\pi r^{2}2sin\frac{\theta }{2}cos\frac{\theta }{2} d\theta$

$\Rightarrow ds=4\pi r^{2}sin\frac{\theta }{2}cos\frac{\theta }{2} d\theta$

Then    $\frac{dF}{ds}=\frac{dN}{N_{0}ds}$

$\Rightarrow \frac{dF}{ds}=\frac{-\pi nta^{2}\frac{cot\frac{\theta }{2}}{sin^{2}\frac{\theta}{2}}d\theta}{N_{0}4\pi r^{2}sin\frac{\theta}{2}.cos\frac{\theta}{2}}$ $\left | \frac {cot\frac{\theta}{2}}{sin^{2}\frac{\theta}{2}}=\frac{\frac{cos\frac{\theta}{2}}{sin\frac{\theta}{2}}}{sin^2\frac{\theta}{2}}=\frac{cos\frac{\theta}{2}}{sin^{3}\frac{\theta}{2}} \right |$

$\Rightarrow \frac{dN}{N_{0}ds}= \frac{-nta^{2}cos\frac{\theta}{2}}{4r^{2}sin^{4}\frac{\theta}{2}.cos\frac{\theta}{2}}$

$\Rightarrow \frac{dN}{N_{0}ds}= \frac{-nta^{2}}{4r^{2}sin^{4}\frac{\theta}{2}}$

$\Rightarrow \frac{dN}{N_{0}ds}= \frac{-nt}{4r^{2}sin^{4}\frac{\theta}{2}}\times \frac{Z^{2}e^{4}}{16\pi ^{2}\varepsilon _{0}^{2}k^{2}}$ $\left ( \therefore a=\frac{Ze^{2}}{4\pi \epsilon _{0}k} \right )$

$\Rightarrow \frac{dN}{ds}=\frac{-N_{0}nt.\: Z^{2}e^{4}}{(8\pi \epsilon _{0})^{2}r^{2}k^{2}sin^{4}\frac{\theta}{2}}$

$\large \Rightarrow\left | \frac{dN}{ds}\right |=N(0)=\frac{N_{0}nt.\: Z^{2}e^{4}}{(8\pi \epsilon _{0})^{2}r^{2}k^{2}sin^{4}\frac{\theta}{2}}$

Where N(0) represents the number of   $\alpha-$ particle per unit area striking the screen. This is the expression for Rutherford's Scattering formula.

Experimental Verification of Rutherford's Alpha particle scattering Formula

The Theory of Rutherford's $\alpha-$ Particle scattering was verified by a series of experiments conducted by Gaiger and Marsden in 1913.

The experimental arrangement consists of an Airtight chamber (c) which can be evacuated by tube (T). The chamber is capable of rotating inside the jacket ( J ) about a vertical axis. Radon is taken as the Radioactive source Rc of $\alpha-$ particles Which is placed inside a Radioactive cell (L).

After coming out of the narrow opening lead cavity the $\alpha-$ particle strikes a thin foil (F) (Gold, Silver, Platinum ).The foil is placed at the centre of the chamber (c).The scattered $\alpha-$ particles are viewed through low Power Microscope (M) which is provided with a fluorescent screen. As the chamber rotates about a central axis the microscope rotates along with it but the cavity (L) and foil(f) remain fixed with the tube. The $\alpha-$ particles striking on the foil get scattered in a different direction. The number of $\alpha-$ particles scattering along different direction can be recorded by observing the scintillations on the fluorescent screen.

The Following Results Obtained From The Above Experiment:-

Most of the $\alpha-$ particle either passed straight through the metal foil or suffered only small deflection. This could be explained by Thomson's Atomic Model.
A few $\alpha-$ particle were deflected through angle where were less than 90 degree ( $\leqslant 90^{0}$ ) and a very few $\alpha-$ particle were deflected through an angle greater than 90 degree ( $> 90^{0}$ ). Sometimes a particle was found to be deflected through 180 degree( $180^{0}$ ) . These large angle scattering could not be explained by Thomson's Atomic Model
If $\large \theta$ is the scattering angle and N is the number of Particles available in that direction then it is found that

$\large \log_{e}N\propto \frac{1}{\sin ^{4}\frac{\theta }{2}}\propto cosec^{4}\frac{\theta }{2}$

which is perfectly agree with the theory.
4.If "t" is the thickness of the foil and N is the number of $\alpha-$ particles scattered along scattering angle $\large \theta$ , It was observed that $N\propto t$ .This agreed with the theory.

5. This experiment also verified that Number of $\alpha-$ particles "N" scattered along scattering angle $\large \theta$ is directly proportional to ( $\propto$ ) square of the Atomic Number of the foil atom i.e $N\propto Z^{2}$

Size of the nucleus from alpha( $\alpha-$ ) particles scattering:-

Let an $\alpha-$ particle having velocity $\vartheta$ approach a nucleus (head on) having a charge (+Ze). The velocity of   $\alpha-$ particle decreases till it comes to rest at a distance "b" from the nucleus.It is the then repelled back along the direction of approach.

Initial Kinetic Energy of   $\alpha-$ particle = $\frac{1}{2}mv^{2}$

Initial Potential Energy of   $\alpha-$ particle = 0

Final Kinetic Energy of   $\alpha-$ particle = 0

Final Kinetic Energy of $\alpha-$ particle =   $\large \frac{1}{4\pi \epsilon _{0}}\, \frac{q_{1}q_{2}}{r_{0}}$

From energy conservation law,

Initial Total Energy = Final Total Energy

$\large \Rightarrow \frac{1}{2}mv^{2}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r_{0}}$

$\large \Rightarrow \frac{1}{2}mv^{2}=\frac{1}{4\pi \epsilon _{0}}\frac{q_{1}q_{2}}{r_{0}}$

$\large \Rightarrow r_{0}= \frac{1}{4\pi \epsilon _{0}}\frac{2q_{1}q_{2}}{mv^{2}}$

the charge of $\alpha-$ particle = 2e
the charge of nucleus = Ze

$\large \Rightarrow r_{0}= \frac{2}{4\pi \epsilon _{0}}.\frac{2e\times Ze}{mv^{2}}= \frac{1}{4\pi \epsilon _{0}}\frac{4Ze^{2}}{mv^{2}}$
$\large \Rightarrow r_{0}= (9\times 10^{9}).\frac{4Z(1.6\times 10^{-19})^{2}}{mv^{2}}$
by putting the value of m,v and Z for given experiment we can find the radius of given nucleus.

Q. In an experiment velocity of   $\alpha-$ particle is   $\large 2\times 10^{7}$ m/s is bombarded upon Gold Z= 79,
Mass of   $\alpha-$ particle M= $\large 4\times 1.67\times 10^{-27} g$ , Then what is the radius of the nucleus?

$\large r_{0}= \frac{4(9\times 10^{9})\times 79\times (1.6\times 10^{-19})^{2}}{4\times (1.67)\times 10^{-27}\times (2\times 10^{7})^{2}}=2.62\times 10^{-14}m$

Failure of Rutherford's scattering formula

1. According to electromagnetic theory, a charged particle in accelerated motion must radiate energy in the form of electromagnetic radiation. As a result, there should be a gradual decrease in the energy of the electron. The electron should follow a spiral path and ultimately fall into the nucleus. thus the whole atomic structure should collapse. this is contrary to the actual fact that atom is very stable.

2. According to Rutherford's mode electron can revolve in any orbit, So it must emit continuous radiation of all frequencies but elements emit spectral lines of only definite frequency.

The Science Thinkers

Nuclear Battery

What Would Happen If You Detonate A Nuclear Bomb In Space?

Rutherford's Alpha particle scattering Experiment

Rutherford's Alpha( $\alpha-$ ) Particle Scattering

Derivation Of Rutherford's Scattering Formula

Experimental Verification of Rutherford's Alpha particle scattering Formula

The Following Results Obtained From The Above Experiment:-

Size of the nucleus from alpha( $\alpha-$ ) particles scattering:-

Failure of Rutherford's scattering formula

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Nuclear Battery

What Would Happen If You Detonate A Nuclear Bomb In Space?

Rutherford's Alpha particle scattering Experiment

Rutherford's Alpha() Particle Scattering

Derivation Of Rutherford's Scattering Formula

Experimental Verification of Rutherford's Alpha particle scattering Formula

The Following Results Obtained From The Above Experiment:-

Size of the nucleus from alpha() particles scattering:-

Failure of Rutherford's scattering formula

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EVEN STAR SYSTEMS HAVE IDENTITY CRISES

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Rutherford's Alpha( $\alpha-$ ) Particle Scattering

Size of the nucleus from alpha( $\alpha-$ ) particles scattering:-