Tag: atomic nuclei

According to Bohr, the wavelength emitted when an electron jumps from ${ n } _{ 1 }^{ th }$ to ${ n } _{ 2 }^{ th }$ orbit is
$E=\dfrac { hc }{ \lambda  } ={ E } _{ 2 }-{ E } _{ 1 }$
$\Rightarrow \dfrac { 1 }{ \lambda  } =R\left( \dfrac { 1 }{ { n } _{ 1 }^{ 2 } } -\dfrac { 1 }{ { n } _{ 2 }^{ 2 } }  \right) $
For first line in Lyman series
$\dfrac { 1 }{ { \lambda  } _{ L } } =R\left( \dfrac { 1 }{ { 1 }^{ 2 } } -\dfrac { 1 }{ { 2 }^{ 2 } }  \right) =\dfrac { 3R }{ 4 } $                        ......(i)
For first line in Balmer series,
$\dfrac { 1 }{ { \lambda  } _{ B } } =R\left( \dfrac { 1 }{ { 2 }^{ 2 } } -\dfrac { 1 }{ { 3 }^{ 2 } }  \right) =\dfrac { 5R }{ 36 } $                      ......(ii)
From equations (i) and (ii)
$\therefore \dfrac { { \lambda  } _{ B } }{ { \lambda  } _{ L } } =\dfrac { 3R }{ 4 } \times \dfrac { 36 }{ 5R } =\dfrac { 27 }{ 5 } $
$\therefore { \lambda  } _{ B }=\dfrac { 27 }{ 5 } \lambda $               $\left( \because { \lambda  } _{ L }=\lambda  \right) $

When transition is from upper state to lower state, the energy is evolved.
Energy is emitted only when an electron jumps from an outer stationary orbit of energy ${ E } _{ 2 }$ to inner stationary orbit of energy ${ E } _{ 1 }$. The energy emitted $E={ E } _{ 2 }-{ E } _{ 1 }$
In given options only (C) and (D) are practicable.
In option (C),
$E=-Rhc\left( \dfrac { 1 }{ { 1 }^{ 2 } } -\dfrac { 1 }{ { 2 }^{ 2 } }  \right) $
     $=-\dfrac { 3 }{ 4 } Rhc$
In option (D),
$E=-Rhc\left( \dfrac { 1 }{ { 2 }^{ 2 } } -\dfrac { 1 }{ { 6 }^{ 2 } }  \right) $
    $=-\dfrac { 2 }{ 9 } Rhc$
Thus, maximum energy is evolved in option (C).
Note: The process in which absorption of energy by an electron takes the electron from an inner orbit to some outer orbit of higher energy is called excitation.

Charles- Augustine de Coulomb was a French physicist known for his work in electrostatics, mainly for developing the Coulomb's law that quantifies the force between two charged particles separated at some distance. 

A major problem with Bohr's model was that it treated electrons as particles that existed in precisely-defined orbits. de Broglie made a different assumption about the electron and its stable orbits which turned out to be mathematically equivalent to Boh'r assumption. He assumed that the electron forms standing wave patterns that must fit a circular shape round the nucleus.

$\lambda=\cfrac{h}{mv}$

de-Broglie wavelength $\lambda=\dfrac{h}{mv}$
Where the speed ($r.m.s$) of a gas particle at the given temperature ($T$) is given as $\dfrac{1}{2} mv^{2}=\dfrac{3}{2}kT$
$\Rightarrow \quad v=\sqrt{\dfrac{3KT}{m}}$ where $k=$ Boltzmanns constant and $m=$ mass of the gas particle and $T=$ temperature of the gas in $k$
$\Rightarrow \quad mv=\sqrt{3mKT}$
$\Rightarrow \quad \lambda=\dfrac{h}{mv}=\dfrac{h}{\sqrt{3mkT}}$
$\therefore \quad \dfrac{\lambda _{H}}{\lambda _{He}}=\sqrt{\dfrac{m _{He}\ T _{He}}{m _{H}\ T _{H}}}$
$=\sqrt{\dfrac{(4\ amu)\ (273 + 127)^{\circ }k}{(2\ amu)\ (273 + 127)^{\circ }k }}=\sqrt{\dfrac{8}{3}}$
Hence ($D$) is correct.

From the formula,

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

Where,

Rydberg Constant,$R=\dfrac{m{{e}^{4}}}{8\varepsilon _{o}^{2}{{h}^{3}}c}$

New Constant, (for $m=2$ ),$\,{{R} _{new}}=\dfrac{\left( 2m \right){{e}^{4}}}{8\varepsilon _{o}^{2}{{h}^{3}}c}=2R$

Longest wavelength,

$ \dfrac{1}{{{\lambda } _{long}}}=2R\left( \dfrac{1}{n _{1}^{2}}-\dfrac{1}{n _{2}^{2}} \right)=2R\left( \dfrac{1}{1}-\dfrac{1}{{{2}^{2}}} \right)=\dfrac{6R}{4} $

$ \Rightarrow {{\lambda } _{long}}=\dfrac{2}{3R} $

Hence, longest wavelength is $\dfrac{2}{3R}$

$J=\dfrac{nh}{2\pi}=mvr$
$J=mvr,\lambda =\dfrac{h}{mv}$
$v\alpha \dfrac{1}{n} ,r\alpha n^2$
$\Rightarrow \lambda \alpha n$
$J\alpha n$
$\Rightarrow J\alpha \lambda$