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".
Average 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.

    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:

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

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

    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.

    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 :
    Emission Transitions of Hydrogen Atom
    (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 $n_{2}= 6,7, 8, ....$

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