Tag: atomic nuclei

Questions Related to atomic nuclei

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 :

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

  1. $(0.529)n\lambda $

  2. $\sqrt { n } \lambda $

  3. $(13.6)\lambda $

  4. $n\lambda $

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Correct Option: D
Explanation:

By De-broglie hypotheseis -

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

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 

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

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

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

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

  4. $\lambda _{p} \propto \lambda _{e}^{2}$

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Correct Option: D
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.

Composition of proton is indicated as ___

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

  1. uud

  2. udd

  3. uuu

  4. ddd

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Correct Option: A

The unit of Planck's constant is equivalent to that of

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

  1. energy

  2. angular momentum

  3. velocity

  4. force

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Correct Option: B
Explanation:

The unit for Planck's constant is $ kg m^{2} s^{-1}$
The SI unit for angular momentum is $ kg m^{2} s^{-1}$
Hence they have the same SI units.
Option B is correct.

If the radius of first Bohrs orbit is $x$, then de-Broglie wavelength of electron in 3rd orbit is nearly

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

  1. $2\pi$ $x$

  2. $6\pi$$x$

  3. $9x$

  4. $x/3$

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Correct Option: B
Explanation:

Radius of 3rd orbit radius $=9x=n^{2}x$ (where $n=3$ )
Let de broglie wavelength be $\lambda $.
For the interference of the waves to be constructive,
$n\lambda =2\pi r$ ($r$ is radius of orbit)
$\Rightarrow \lambda =\dfrac{2\pi \times 9x}{3} $ (where, $\ n=3 $, the quantum state)
$\Rightarrow \lambda =6\pi x$

The circumference of the second orbit of an atom or ion having single electron ,is $4 \times10^{-9}$ m.The de-Brogile wavelength of electron revolving in this orbit should be

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

  1. $2\times 10^{-9}m$

  2. $4\times 10^{-9}m$

  3. $8\times 10^{-9}m$

  4. $1\times 10^{-9}m$

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Correct Option: A
Explanation:

The circumference of the orbit $=4\times 10^{9}m$
The orbit number $=2$

$n\lambda =2\pi r$

$\Rightarrow \lambda =\dfrac{4\times 10^{9}}{2}m$

$\Rightarrow \lambda =2\times 10^{-9}m$

If the electron in hydrogen orbit jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by

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

  1. $\lambda = \dfrac {R}{6}$

  2. $\lambda = \dfrac {5}{R}$

  3. $\lambda = \dfrac {36}{5R}$

  4. $\lambda = \dfrac {5R}{36}$

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Correct Option: C
Explanation:

We know that
$\dfrac {1}{\lambda} = R \left (\dfrac {1}{n _{1}^{2}} - \dfrac {1}{n _{2}^{2}} \right )$
$\dfrac {1}{\lambda} = R\left (\dfrac {1}{2^{2}} - \dfrac {1}{3^{2}} \right ) \Rightarrow R \left (\dfrac {1}{4} - \dfrac {1}{9}\right )$
$\dfrac {1}{\lambda} = \left (\dfrac {9 - 4}{36}\right ) R = \dfrac {5R}{36} \Rightarrow \lambda = \dfrac {36}{5R}$

According to de-Broglie explanation of Bohr's second postulate of quantization, the standing particle wave on a circular orbit for $n = 4$ is given by

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

  1. $2 \pi {r} _{n} = {4}/{\lambda}$

  2. $\dfrac{2 \pi}{\lambda} = 4{r} _{n}$

  3. $2 \pi {r} _{n} = 4 \lambda$

  4. $\dfrac{\lambda}{2 \pi} = 4 {r} _{n}$

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Correct Option: C
Explanation:

According to debroglie explanation of Bohr's second postulate, assumption is made that integral number of wavelengths must fir in the circumference of circular orbit. The integral multiple comes out to be the same as quantization number.

$2 \pi r _n = n \lambda$
For $n=4$, 
       $2 \pi r _n = 4 \lambda$

Which of the following relations is correct?

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $E = mc$

  2. $E = mc^2$

  3. $E = 2mc^2$

  4. $E = mc^2/4$

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Correct Option: B
Explanation:

According to Einstein, energy and mass are related by the relation.
$E = mc^2$
where c is the speed of light in vacuum.

One milligram of matter is converted into energy. The energy released will be

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $9\times 10^{6} J$

  2. $9\times 10^{8}J$

  3. $9\times 10^{10}J$

  4. $9\times 10^{12}J$

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Correct Option: C
Explanation:

Here, $m = 1\ mg = 1\times 10^{-3} g $

             $= 1\times 10^{-6}kg$
According to Einstein mass-energy equivalence
                $E = mc^{2}$
where $c$ is the speed of light in vacuum
$\therefore E = (1\times 10^{-6} kg)(3\times 10^{8}ms^{-1})^{2}$
$ = 9\times 10^{10}J$.