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Atoms Test - 68

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Atoms Test - 68
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  • Question 1
    1 / -0
    Monochromatic radiations of wavelength $$\lambda$$ are incident on a hydrogen sample in the ground state. Hydrogen atom absorbs the light and subsequently emits radiations of $$10$$ different wavelengths. The value of $$\lambda$$ is nearly
    Solution
    Total no. of wavelengths emitted $$=\ ^n C _2= \cfrac{n(n-1)}{2}$$
    $$^n C _2 = 10$$
    $$ n^2 -n -20 =0$$
    $$ n=-4 \ or \:5$$
    $$n=5$$

    So the energy required:
    $$ \cfrac{1}{\lambda} = R \left (1 - \cfrac{1}{5^2}\right )$$
    $$ R = 1.1 \times 10^7$$
    $$ \lambda = 94.6\ nm$$

  • Question 2
    1 / -0
    If elements of quantum number greater than $$'n'$$ were not allowed, the number of possible elements in nature would have been
    Solution
    $${ n }^{ th }$$ shell can hold upto $$2{ n }^{ 2 }$$ electrons
    Total number of possible elements would be $$\sum _{ 1 }^{ n }{ 2{ n }^{ 2 } } $$$$=\dfrac { n }{ 3 } \left( n+1 \right) \left( 2n+1 \right) $$
  • Question 3
    1 / -0
    If a hydrogen atom emits a photon of energy $$12.1\space eV$$, its orbital angular momentum changes by $$\triangle L$$. Then, $$\triangle L$$ equals
    Solution
    Using the attached formula:
    $$ 12.1 = 13.6 \times (\cfrac{1}{n_1 ^2} - \cfrac{1}{n_2 ^2})$$
    $$(\cfrac{1}{n_1 ^2} - \cfrac{1}{n_2 ^2}) = \cfrac{8}{9}$$

    Hence, $$n_1 =1$$ and $$n_2 = 3$$
    Change in Angular momentum $$= (n_2 -n_1) \cfrac{h}{2 \pi}$$$$= 2.11 \times 10^{-34}  J$$
  • Question 4
    1 / -0
    If potential energy between a proton and an electron is given by $$|U| = ke^2/2R^3$$, where $$e$$ is the charge of electron and $$R$$ is the radius of atom, then radius of Bohr's orbit is given by ($$h$$ = Planck's constant, $$k$$ = constant)
    Solution
    Potential energy at a distance $$R$$ is given as, $$U=\dfrac{ke^2}{2R^3}$$
    Thus force between proton and electron $$=-\dfrac{dU}{dR}$$ $$=\dfrac{3ke^2}{2R^4}$$
    This force causes the centrifugal acceleration of the system.
    Thus, $$\dfrac{3ke^2}{2R^4}=\dfrac{mv^2}{R}$$
    Bohr's postulate is: $$mvR=\dfrac{nh}{2\pi}$$
    Substituting the expression for $$v$$ from above equation gives,
    $$\dfrac{3ke^2}{2R^4}=\dfrac{m}{R}(\dfrac{nh}{2\pi m R})^2$$

    $$\implies R=\dfrac{6ke^2\pi^2m}{n^2h^2}$$
  • Question 5
    1 / -0
    The wavelength of $$K_\alpha$$ X-rays of two metals $$A$$ and $$B$$ are $$\dfrac{4}{1875R}$$ and $$\dfrac{1}{675R}$$, respectively, where $$\mbox{'R'}$$ is Rydberg's constant. The number of elements lying between $$A$$ and $$B$$ according to their atomic number is
    Solution
    In X-ray spectroscopy, K-alpha emission lines result when an electron transitions to the innermost "K" shell (principal quantum number 1) from a 2p orbital of the second or "L" shell (with principal quantum number 2). 
     $$ \cfrac{1}{\lambda}= Z^2 R(1 - \cfrac{1}{2^2}) = \cfrac{3RZ^2}{4}$$
    $$ \lambda _A =  \cfrac{4}{1875R} = \cfrac{4}{3 R Z^2}$$
    $$ Z_A =25$$

    Similarly,
    $$ \lambda _B =  \cfrac{1}{675R} = \cfrac{4}{3 R Z_B ^2}$$
    $$Z_B = 30$$

    No. of elements between A and B $$= 4$$
  • Question 6
    1 / -0

    Directions For Questions

    In an ordinary atom, as a first approximation, the motion of the nucleus can be ignored. In a positronium atom, a positron replaces the proton of hydrogen atom. The electron and positron masses are equal and therefore, the motion of the positron cannot be ignored. One must consider the motion of both electron and positron about their center of mass. A detailed analysis shows that formulae of Bohr's model apply to positronium atom provided that we replace $$m_e$$ by what is known as the reduced mass of the electron. For positronium, the reduced mass is $$m_e/2$$.

    ...view full instructions

    The orbital radius of the first excited level of positronium atom is

    Note: $$a_0$$ is the orbital radius of ground state of hydrogen atom
    Solution
    Given, electron in the positronium has reduced mass $$\dfrac { { m }_{ e } }{ 2 }$$
    And orbital radius of ground state of hydrogen atom is $${ a }_{ 0 }$$.
    Therefore, for positronium, $${ r }_{ n }=2{ a }_{ 0 }{ n }^{ 2 }$$
    Hence, orbital radius of first excited level of positronium  is $$8{ a }_{ 0 }$$.
  • Question 7
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    Directions For Questions

    The electrons in a $$H$$-atom kept at rest, jumps from the $$mth$$ shell to the $$nth$$ shell $$(m>n)$$. Suppose instead of emitting electromagnetic wave, the energy released is converted into kinetic energy of the atom. Assume Bohr's model and conservation of angular momentum are valid. Answer the following question

    ...view full instructions

    What principle is violated here?
    Solution
    Laws of Motion:
    Newtons second law: $$F = ma$$
    Without any force, the atom cannot accelerate.
    Conservation of Momentum:
    The linear momentum of the atom can't change without any external force acting on it.
  • Question 8
    1 / -0

    Directions For Questions

    An electron orbits a stationary nucleus of charge $$+ze$$, where $$z$$ is a constant and $$e$$ is the magnitude of electronic charge. It requires $$47.2\space eV$$ to excite the electron from the second Bohr orbit to the third Bohr Orbit.

    ...view full instructions

    The radius of the first Bohr orbit is
    Solution
    Given, $$47.2\ eV$$ is required to excite an electron from $$n=2$$ to $$n=3$$
    $$\Rightarrow  \dfrac { 13.6 { Z }^{ 2 } }{ 4 } -\dfrac { 13.6 { Z }^{ 2 } }{ 9 } =\dfrac { 5\times 13.6 { Z }^{ 2 } }{ 36 } =47.2$$ 

    $$\therefore \ { Z }^{ 2 }= 24.99$$

    Therefore, $$Z=5$$

    Radius of first Bohr orbit is $$=0.529\times \dfrac { { 1 }^{ 2 } }{ 5 } =0.1058\mathring { A }$$

    $$\Rightarrow \ { r }_{ 1 }= 0.1058\times { 10 }^{ -10 }m$$
  • Question 9
    1 / -0

    Directions For Questions

    Hydrogen is the simplest atom of nature. There is one proton in its nucleus and an electron moves around the nucleus in a circular orbit. According to Niels Bohr, this electron moves in a stationary orbit. When this electron is in the stationary orbit, it emits no electromagnetic radiation. The angular momentum of the electron is quantized, i.e., $$mvr = (nh/2\pi)$$, where $$m$$ = mass of the electron, $$v$$ = velocity of the electron in the orbit, $$r$$ = radius of the orbit, and $$n = 1,2,3,...$$. When transition takes place from the $$Kth$$ orbit to the $$Jth$$ orbit, energy photon is emitted. If the wavelength of the emitted photon is $$\lambda$$, we find that $$\displaystyle\frac{1}{\lambda} = R\left[\displaystyle\frac{1}{J^2} - \displaystyle\frac{1}{K^2}\right]$$, where $$R$$ is the Rydberg's constant. On a different planet, the hydrogen atom's structure was somewhat different from ours. The angular momentum of electron was $$P = 2n(h/2\pi)$$, i.e., an even multiple of $$(h/2\pi)$$.

    Answer the following questions regarding the other planet based on the above passage.

    ...view full instructions

    The minimum permissible radius of the orbit will be :

    Solution
    Minimum radius will be when $$n=1$$:
    $$mvR_\circ{}=2 \dfrac{h}{2\pi}\Rightarrow vR_\circ{}= \dfrac{h}{m\pi}$$...(i)

    Electrostatic force is providing centripetal force for electron's circular motion:
    $$\dfrac{1}{4\pi\epsilon_\circ{}}\dfrac{e^2}{R_\circ{}^2}=\dfrac{mv^2}{R_\circ{}}$$...(ii)

    $$\dfrac{1}{4\pi\epsilon_\circ{}}\dfrac{e^2}{m}=v^2R_\circ{}$$...(iii)

    Substituting $$vR_\circ{}$$ from equation (i)
    $$v \dfrac{h}{m\pi}=\dfrac{1}{4\pi\epsilon_\circ{}}\dfrac{e^2}{m}$$
    $$v =\dfrac{e^2}{4\epsilon_\circ{}h}$$

    From relation (i) $$R_\circ{}=\dfrac{h}{m\pi v}$$
    $$ R_\circ{}=\dfrac{h^2 4\epsilon_\circ{}}{me^2\pi}$$
  • Question 10
    1 / -0

    Directions For Questions

    A sample of hydrogen gas in its ground state is irradiated with photons of $$10.2\space eV$$ energies. The radiation from the above sample is used to irradiate two other samples of excited ionized $$He^+$$ and excited ionized $$Li^{2+}$$, respectively. Both the ionized samples absorb the incident radiation.

    ...view full instructions

    How many spectral lines are observed in spectra of $$He^+$$ ion?
    Solution
    Energy levels of helium are: $$-54.4eV, -13.6eV, -6.04eV$$ and so on.
    Given, helium ion electron is in excited state. 
    Occurrence of spectral lines is possible only if it is in $$n=2$$, when given $$10.2\ eV$$, because if the electron is in any other level, atom ionizes.
    $${ E }_{ 4 }=-3.4\ eV$$
    Therefore, electron excites into $$n=4$$.
    No. of emission lines is $$\dfrac { 4\times 3 }{ 2 } =6$$
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