Tag: atomic nuclei

Quizzes (8) Questions (121)

Questions Related to atomic nuclei

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.

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

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

  1. $2\pi$ $x$
  2. $6\pi$$x$
  3. $9x$
  4. $x/3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)
$\Rightarrow \lambda =6\pi x$

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

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

  1. $2\times 10^{-9}m$
  2. $4\times 10^{-9}m$
  3. $8\times 10^{-9}m$
  4. $1\times 10^{-9}m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

$n\lambda =2\pi r$

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

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

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

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

  1. $\lambda = \dfrac {R}{6}$
  2. $\lambda = \dfrac {5}{R}$
  3. $\lambda = \dfrac {36}{5R}$
  4. $\lambda = \dfrac {5R}{36}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  1. $2 \pi {r} _{n} = {4}/{\lambda}$
  2. $\dfrac{2 \pi}{\lambda} = 4{r} _{n}$
  3. $2 \pi {r} _{n} = 4 \lambda$
  4. $\dfrac{\lambda}{2 \pi} = 4 {r} _{n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $n=4$, 
       $2 \pi r _n = 4 \lambda$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

Which of the following relations is correct?

  1. $E = mc$
  2. $E = mc^2$
  3. $E = 2mc^2$
  4. $E = mc^2/4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

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

  1. $9\times 10^{6} J$
  2. $9\times 10^{8}J$
  3. $9\times 10^{10}J$
  4. $9\times 10^{12}J$
Reveal answer Fill a bubble to check yourself
C Correct answer
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$.