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Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Maximum energy is evolved during which of the following transitions?

  1. $n=1$ to $n=2$

  2. $n=2$ to $n=6$

  3. $n=2$ to $n=1$

  4. $n=6$ to $n=2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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.

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Which of the following option is NOT a correct pairing of physicist and field in which he made significant contributions?

  1. Newton - gravitation

  2. Einstein - relativity

  3. Faraday - electricity and magnetism

  4. Coulomb - quantum mechanics

  5. Bohr - atomic structure
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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. 

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

One of Bohr's assumptions about stable electron orbits in a hydrogen atom treated electrons as particles with circular orbits. Louis de Broglie made a different assumption about the electron and its stable orbits which turned out to be mathematically equivalent to Bohr' assumption.
What assumption did the Broglie make?

  1. The electron orbits the nucleus with elliptical-shaped orbits, like the planets around the sun

  2. The electron forms standing wave patterns that must fit a circular shape around the nucleus

  3. The electron follows a parabolic trajectory around the nucleus

  4. The electron behaves like a cloud, blocking out all radiation except radiation associated with allowable energy transitions

  5. The electron follows a hyperbolic trajectory around the nucleus
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Why are matter waves significant only in the motion of extremely small particles?

  1. Small particles have smaller mass and hence the de Broglie wavelength increases

  2. Small particles are manageable easily

  3. Small particles possess small kinetic energy

  4. Extremely small particles possess small relativistic mass
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Smaller particles have larger wavelength and more wave like character.

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

The ratio of de-Broglie wavelength of molecules of hydrogen and helium in two gas jars kept separately at temperature $27^{o}C$ and $127^{o}C$ respectively is

  1. $\dfrac{2}{\sqrt{3}}$

  2. $2:3$

  3. $\dfrac{\sqrt{3}}{4}$

  4. $\sqrt{\dfrac{8}{3}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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.

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Imagine an atom made of a proton and a hypothetical partical of double the mass as that of an electron but the same charge. Apply Bohr theory to consider transitions of the hypothetical particle to the ground state. Then, the longest wavelength (in terms of Rydberge constant for hydrogen atom) is

  1. $\dfrac {1}{2R}$

  2. $\dfrac {5}{3R}$

  3. $\dfrac {1}{3R}$

  4. $\dfrac {2}{3R}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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 $\dfrac{2}{3R}$

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

de - Broglie wavelength of an electron in the nth bohr orbit is $\lambda _n$ and the angular momentum is $J _n$ then:

  1. $J _n\alpha \lambda _n$

  2. $\lambda _n \infty \dfrac{1}{J _n}$

  3. $\lambda _n \infty J _n^2$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

Consider an electron in the ${ n }^{ th }$ orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de Brogile wavelength ${ n }^{ th }$ of the electron as :

  1. $(0.529)n\lambda $

  2. $\sqrt { n } \lambda $

  3. $(13.6)\lambda $

  4. $n\lambda $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By De-broglie hypotheseis -

$\begin{array}{l} mvr=\dfrac { { nh } }{ { 2\pi  } }  \ 2\pi r=n\left( { \dfrac { h }{ { mv } }  } \right) =n\lambda  \end{array}$

Multiple choice limitations of bohr model and explanation of bohr's second postulate by matter waves bohr's model atoms atomic nuclei physics

If photon energy $E$ and an electron have same energy $E$ (kinetic energy) and De-Broglie wavelength of an electron is $\lambda _{e}$ and De-Broglie wavelength of photon is $\lambda _{p}$. The correct relation between $\lambda _{e}$ and $\lambda _{p}$ is 

  1. $\lambda _{p} \propto \lambda _{e}$

  2. $\lambda _{p} \propto \sqrt{\lambda _{e}}$

  3. $\lambda _{p} \propto \dfrac{1}{\sqrt{\lambda _{e}}}$

  4. $\lambda _{p} \propto \lambda _{e}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For photon

$E = \dfrac{{hc}}{{{\lambda _p}}}$---------------$(1)$
$ \Rightarrow {\lambda _p} = \dfrac{{hc}}{E}$

$E = \dfrac{1}{2}m{v^2}$

$ \Rightarrow v = \sqrt {\dfrac{{2E}}{m}} $

${\lambda _e} = \dfrac{h}{{mv}} = \dfrac{h}{{n\sqrt {\dfrac{{2E}}{m}} }} = \frac{h}{{\sqrt {2Em} }}$

$ \Rightarrow {\lambda _e}^2 = \dfrac{{{h^2}}}{{2Em}}$

$ \Rightarrow E = \dfrac{{{h^2}}}{{2m{\lambda _e}^2}}$
$\therefore \dfrac{{{h _e}}}{{{\lambda _p}}} = \dfrac{{{h^2}}}{{2m{\lambda _e}^2}}$

$ \Rightarrow {\lambda _p}\alpha \,{\lambda _e}^2$
Hence,
option $(D)$ is correct answer.