Binding Energy Curve

Binding Energy Curve :

A graph is plotted for different nuclei between the binding energy per nucleon and the atomic mass number. This graph gives a curve which is called " binding energy curve".

There are following discussion point obtained from the binding energy curve :

1.) For Nuclei with $A=50$ TO $A=80$:

For nuclei with atomic mass number $A = 50 - 80$ , the B.E./ nucleon (i.e. binding energy per nucleon) is approximately $8.5 MeV$.

The curve is almost flat in this and indicate the highly stability of the nucleus.

2.) For Nuclei with $A \geq 80$:

For heavier nuclei with $A \gt 80$, the B.E. /nucleon ( i.e. binding energy per nucleon) decreases slowly and reaching about $7.6 MeV$ for uranium ($U \: A = 238$).

The lower value of binding energy per nucleon fails to counteract the Coulombian repulsion among protons in nuclei having large number of protons resulting instability

Consequently, the nuclei of heavier atoms beyond $_{83}Bi^{209}$ are radioactive.

3.) For Nuclei with $A \leq 50$:

For nuclei with atomic mass number below $50$ , the B.E./ nucleon decreases, with a sharp drop below $A=20$.

For example: Heavy hydrogen (i.e $_{1}H^{2}$), it is only about $1.1 MeV$. it indicates that lower stability for nuclear with mass number below $20$.

4.) Subsidiary Peak for $A \lt 50$:

Below $A = 50$, the curve does not fall continuously, but the subsidiary peaks at $_{8}O^{16}, _{6}C^{12},_{2}He^{4}$.

These peak indicate that such even-even nuclear are more stable compared to the immediate neighbours .

5.) Nuclear fusion and Nuclear fission process release energy:

From curve, it shows that drops down in curve at both high and low mass number and lower binding energy per nucleon.

For example:

A very high amount of energy is released in the process of nuclear fission and fusion because of Lo binding energy causes instability of the nucleus.

Read More

Nuclear Fission and Nuclear Fusion

Nuclear Fission:

When a heavy nucleus breaks into two or more smaller, lighter nuclei and produces high energy, this process is called as nuclear fission.

Example:

$_{92}U^{235} +\: _{0}n^{1} (Neutron) \rightarrow \: _{92}U^{236} \rightarrow _{56}Ba^{141} + \: _{36}Kr^{92} + \: 3 _{0}n^{1} + \gamma$

Nuclear Fusion:

When two or more very light nuclei move with a very high speed then these nuclei are fused and form a single nucleus. This process is called as nuclear fusion.

Example: Two deuterons can be fused to form a triton(tritium nucleus) as shown in the reaction below:

$_{1}H^{2} + \: _{1}H^{2} \rightarrow \: _{1}H^{3} + \: _{1}H^{1} + \: 4.0 \: MeV \:(Energy)$

$_{1}H^{3} (Tritium) + _{1}H^{2} \rightarrow \: _{2}He^{4} + _{0}n^{1} + 17.6.0 \: MeV \:(Energy)$

The total result of the above two equations is the fusion of deuterons and produces an $\alpha - $ particle $(_{2}He^{4})$, a neutron $(_{0}n^{1})$ and a proton $(_{1}H^{1})$. The total released energy is $21.6 MeV$.

Alternatively, the fusion of three deutrons $(_{1}H^{2})$ into $\alpha -$ partice can takes place as follows:

$_{1}H^{2} + _{1}H^{2} \rightarrow \: _{2}He^{3} + _{0}n^{1} + 3.3 \: MeV \:(Energy)$

$_{2}He^{3} + _{1}H^{2} \rightarrow \: _{2}He^{4} + _{1}H^{1} + 18.3 \: MeV \:(Energy)$

Read More

Mass Defect, Binding Energy and Binding Energy per nucleon

Binding Energy:

The difference between the total mass of individual nucleons (i.e. total number of proton and neutron) and actual mass of nucleus of that energy is called binding energy.

$\Delta m = \left (P \times m_{P} + N \times m_{N} \right) - m_{actual} \qquad (1)$

Where
$\Delta m \rightarrow$ Mass Defect
$P \rightarrow$ Number of Proton
$N \rightarrow$ Number of Neutron
$m_{actual} \rightarrow$ Actual mass of nucleus
$m_{P} \rightarrow$ Mass of a Proton
$m_{N} \rightarrow$ Mass of a Neutron

We know that

$Z=P=e \\ N=A-Z \qquad (2)$

Where
$Z \rightarrow $ Atomic Number
$A \rightarrow $ Atomic Mass Number
$ e \rightarrow $ Number of Electrons

From above two equation $(1)$ and equation $(2)$

$\Delta m = \left [ Z \times m_{P} + \left ( A-Z \right) \times m_{N} \right] - m_{actual} \qquad (1)$

Binding Energy:

The energy require to form or break a nucleous is called the binding energy of nucleous.

$B.E= \Delta m \times c^{2} Joule$

Where $B.E.\rightarrow$ Binding Energy

$B.E= \Delta m (in \: a.m.u.) \times 931.5 \: MeV$

Where $1 \: a.m.u. = 1.67377 \times 10^{-27} kilograms$

Binding energy per nucleon:

The energy require to emit one nucleon from the nucleous is called binding energy per nucleon.

$B.E. \: per \: nucleon = \frac{B.E.}{ Total \: No. \: of \: Nucleons}$

Where $B.E.\rightarrow$ Binding Energy

Note: Higher binding energy per nucleon shows higher stability of the nucleus.

Read More

Spectrum of Hydrogen Atom

Description: The different series of hydrogen spectra can be explained by Bohr's theory. According to Bohr's theory, If the ionized state of a hydrogen atom be taken zero energy level, then energies of different energy levels of the atom can be expressed by following the formula

$E_{n}=\frac{Rhc}{n^{2}} \qquad (1)$

Where
$R \rightarrow$ Rydberg's Constant
$h \rightarrow$ Planck's Constant
$n \rightarrow$ Quantum Number

According to Plank's Theory

$E_{2} - E_{1} =h \nu \qquad(2)$

So from equation $(1)$

$E_{1}=\frac{Rhc}{n^{2}_{1}} $ and $E_{2}=\frac{Rhc}{n^{2}_{2}} \qquad (3)$

From equation $(2)$ and equation $(3)$

$\frac{Rhc}{n^{2}_{2}} - \frac{Rhc}{n^{2}_{1}} =h \nu $

$\frac{Rhc}{n^{2}_{2}} - \frac{Rhc}{n^{2}_{1}} = \frac{hc}{\lambda} $

$\frac{1}{\lambda}=R \left(\frac{1}{n^{2}_{1}} -\frac{1}{n^{2}_{2}} \right)$

The quantity $\frac{1}{\lambda}$ is called the 'wave number', All the series found in the hydrogen spectrum are explained by the above equation :

(i) Lyman Series: When an atom comes down from some higher energy level (i.e. $n_{2} = 2, 3, 4, ...$) to the first energy level (lowest energy level), (i.e. $n_{1}= 1$), then spectral lines are emitted in the spectrum region of ultraviolet. The equation for obtaining the wavelengths of these spectral lines:

$\frac{1}{\lambda}=R \left(\frac{1}{1^{2}} -\frac{1}{n^{2}_{2}} \right)$

Where $n_{2} = 2, 3, 4, ...$

In 1916, Lyman photographed the lines of this series of hydrogen spectra. Hence, this series is named Lyman series'. The longest wavelength of this series (for $n_{2} = 2$) is $1216 A^{\circ}$ and the shortest wavelength (for $n_{2} = \infty$) is $912 A^{\circ}$. The wavelength $912 A^{\circ}$ corresponding to $n = \infty$ is called the 'series limit'.

(ii) Balmer Series: When an atom comes down from some higher energy level (i.e. $n_{2} = 3, 4, 5, ...$) to the second energy level (i.e. $n_{1}= 2$), then the spectral lines are emitted in the spectrum region of the visible part.

$\frac{1}{\lambda}=R \left(\frac{1}{2^{2}} -\frac{1}{n^{2}_{2}} \right)$

where $n_{2} = 3, 4, 5, ...$

In 1885, Balmer saw and studied first time these spectral lines. The longest wavelength of this series (for $n_{2} = 3$) is $6563 Å$ and the shortest wavelength (for $n_{2} = \infty$) is 3646 Ä.

(iii) Paschen Series: When an atom comes down from some higher energy level (i.e. $n_{2} = 3, 4, 5, ...$) to the third energy level (i.e. $n_{1}= 3$) then the spectral lines are emitted in the spectrum region of infrared.

$\frac{1}{\lambda}=R \left(\frac{1}{3^{2}} -\frac{1}{n^{2}_{2}} \right)$

where $n_{2} = 4, 5, 6, ...$

(iv) Brackett Series: When an atom comes down from some higher energy level (i.e. $n_{2} = 5, 6, 7, ...$) to the fourth energy level (i.e. $n_{1}= 4$), then the spectral lines are also emitted in the spectrum region of infrared.

$\frac{1}{\lambda}=R \left(\frac{1}{4^{2}} -\frac{1}{n^{2}_{2}} \right)$

where $n_{2} = 5,6, 7,.....$

(iv) Pfund Series: When an atom comes down from some higher energy level (i.e. $n_{2} = 6, 7, 8, ...$) to the fifth energy level (i.e. $n_{1}= 5$) then the spectral lines are also emitted in the spectrum region of infrared.

$\frac{1}{\lambda}=R \left(\frac{1}{5^{2}} -\frac{1}{n^{2}_{2}} \right)$

where $n_{2}= 6,7, 8, ....$

Read More

Radioactive Decay and its types

Definition:

When the unstable atom (called radionuclide) loses its energy through ionizing radiation, this process is known as radioactive decay.

Types of radioactive decay:

There are 3- types of radioactive decay

1. Alpha Decay
2. Beta Decay
3. Gamma Decay

1. Alpha Decay: A helium nuclei which contain two protons and two neutrons is known as an alpha particle. The $\alpha$- particles are commonly emitted by the heavier radioactive nuclei. When the $\alpha$- particle is emitted from the nucleus then the atomic number is reduced by two (i.e. $Z-2$) or the atomic mass number is reduced by 4 (i.e. $A-4$).

Example:

The decay of $Pu^{239}$ into fissionable $U^{235}$ by the emission of $alpha$- particle

$_{94}Pu^{214} \rightarrow _{92}U^{235} + _{2}He^{4} \left(\alpha - particle \right)$

2. Beta Decay: The emission of $\beta$-particle occurs due to the conversion of a neutron into a proton or vice versa in the nucleus. The $\beta$-decay is commonly accompanied by the emission of neutrino ($\nu$) radiation. There are two types of $\beta$-decay.

i.) Beta Minus: When a neutron is converted into a proton then an electron ($_{-1}e^{\circ}$) i.e.$\beta$-minus particle is emitted. When the $\beta$- minus particle is emitted from the nucleus then the atomic number is increased by one (i.e. $Z+1$) and no change in atomic mass number ($A$).

Example:

$_{6}C^{14} \rightarrow _{7}N^{14} + _{-1}e^{\circ} + \overline{\nu}_{e} \: (anti\:neutrino)$

ii.) Beta Plus: When a proton is converted into a neutron then a positron ($_{+1}e^{\circ}$) $\beta$- plus partice is emitted. When the $\beta$- plus particle is emitted from the nucleus then the atomic number is decreased by one (i.e. $Z-1$) and no change in atomic mass number ($A$). It is also known as positron decay. Positron decay is caused when the radioactive nucleus contains an excess of protons.

Example:

$_{12}Mg^{23} \rightarrow _{11}Na^{23} + _{+1}e^{\circ} + \nu_{e}\: (neutrino)$

The penetrating power of $_{-1}\beta^{\circ}$ particles is small compared to $\gamma$-rays, however it is larger than that of $\alpha$-particles.

Note:

Electron Capture: The nucleus captures the electron from orbits and combines with a proton to form a neutron and emits a neutrino.

Example:

$_{26}Fe^{55} + _{-1}e^{\circ} \rightarrow _{25}Mn^{55} + \nu_{e}\: (neutrino)$

3. Gamma (y) Decay: $\gamma$-particles are electromagnetic radiation of extremely short wavelength and high frequency resulting in high energy. The $\gamma$-rays originate from the nucleus while X-rays come from the atom. $\gamma$-wavelength are on average, about one-tenth those of X-rays, though energy ranges overlap somewhat. There is no alternation of atomic or mass numbers due to $\gamma$ decay.

Example:

$_{27}Co^{60} \rightarrow _{27}Co^{60} + \gamma \: (gamma)$

Read More

Combination of cell in the circuit

A.) Combination of cells when emf of cells are same: There are three types of combinations of cells in the circuit

1.) Series Combination of Cells

2.) Parallel Combination of Cells

3.) Mixed Combination of Cells

1.) Series Combination of Cells: Let us consider that the $n$ - cells having emf (electromotive force) $E$ and internal resistance $r$ are connected in series with external resistance $R$. Then from the figure given below

The total emf of the $n$ - cell = $nE$

The total internal resistance of the $n$ - cell = $nr$

The total resistance of the circuit = $nr+R$

The total current in the circuit

$i=\frac{Total \: emf \: of \: the \: n - series \: cell}{Total \: resistance \: of \: the \: circuit}$

$i=\frac{nE}{nr+R}$

2.) Parallel Combination of Cells: Let us consider that the $n$ - cells having emf (electromotive force) $E$ and internal resistance $r$ are connected in parallel with external resistance $R$. Then from the figure given below

The total emf of the $n$ - cell = $E$

The total internal resistance of the $n$ - cell

$\frac{1}{r_{eq}} = \frac{1}{r}+ \frac{1}{r}+.........n \: times$

$\frac{1}{r_{eq}}=\frac{n}{r}$

$r_{eq}=\frac{r}{n}$

The total resistance of the circuit = $\frac{r}{n}+R$

The total current in the circuit

$i=\frac{Total \: emf \: of \: the \: n - parallel \: cell}{Total \: resistance \: of \: the \: circuit}$

$i=\frac{E}{\frac{r}{n}+R}$

$i=\frac{E}{\frac{r+nR}{n}}$

$i=\frac{nE}{r+nR}$

3.) Mixed Combination of Cells: Let us consider that the $n$ - cells having emf (electromotive force) $E$ and internal resistance $r$ are connected in series in each row of $m$ parallel rows with external resistance $R$. Then from the figure given below

The total emf of the $n$ - cell in each row of $m$ parallel rows of the cells = $nE$

The internal resistance of the $n$ - cell in each row = $nr$

The total internal resistance of the $n$ - cell in each of $m$ parallel rows of the cells = $nr$

$\frac{1}{r_{eq}} = \frac{1}{nr}+ \frac{1}{nr}+.........m \: times$

$\frac{1}{r_{eq}}=\frac{m}{nr}$

$r_{eq}=\frac{nr}{m}$

The total resistance of the circuit = $\frac{nr}{m}+R$

The total current in the circuit

$i=\frac{Total \: emf \: of \: the \: cell}{Total \: resistance \: of \: the \: circuit}$

$i=\frac{nE}{\frac{nr}{m}+R}$

$i=\frac{nE}{\frac{nr+mR}{m}}$

$i=\frac{mnE}{nr+mR}$

It is clear from the above equation that for the value of $i$ to be maximum, the value of $(nr+mR)$ should be minimum. Now,

$nr+mR= \left[ \sqrt{nr}-\sqrt{mr} \right]^{2}+2 \sqrt{mnRr}$

Therefore, for $(nr+mR)$ to be minimum, the quantity $\left[ \sqrt{nr}-\sqrt{mr} \right]^{2}$ should be minimum. So

$\left[ \sqrt{nr}-\sqrt{mr} \right]^{2} = 0$

$ \sqrt{nr}-\sqrt{mr} = 0$

$ \sqrt{nr} = \sqrt{mr} $

$nr=mR$

$R=\frac{nr}{m}$

Here, $\frac{nr}{m}$ is the total resistance of the cells.

Thus, When the total internal resistance of the cells are equal to the external resistance then the total current in the external circuit will be maximum in the mixed combination of cells.

Read More