MCQ Questions for Class 12 Medical Physics Atoms Quiz 12

In the figure six lines of emission spectrum are shown. Which of them will be absent in the absorbtion spectrum.

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1,2,3
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1,4,6
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4,5,6
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1,2,3,4,5,6

Explanation

Here all electrons are coming from the higher state to lower state. So in the absorption spectrum there should be lower to higher state.
So,

$1,2,3,4,5,6$ are the absent in the absorption spectrum.

According to drop model of nucleus which of the following cannot be explained ?

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Fission
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Fusion
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$\alpha-$ spectrum
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All of these

A particle moving with a velocity

$\dfrac{1}{10}th$ of light will cross a nucleus in about:

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$10^{-8}sec$
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$10^{-12}sec$
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$10^{-47}sec$
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$10^{-21}sec$

Stability of half filled sub-shell is caused by :

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exchange energy
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greater spin above
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both $(a)$ and $(b)$
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none of the above

Who proved that atoms had a nucleus?

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Ernest Rutherford
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James Chadwick
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J.J. Thomson
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None of these

The mass of an electron in motion depends upon :

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direction fotion
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its velocity
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initiial mass of $e^-$
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its shell number

In Bohr's radius is

$R_0$ then radius of

$3^{rd}$ orbit of hydrogen atom will be :

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$3R_0$
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$6R_0$
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$9R_0$
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$12R_0$

Mark out the incorrect statement:

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A free neutron can transform itself into photon
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A free proton can transform itself into neutron
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In beta minus decay, the electron originates from nucleus
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All of the above

Explanation

When a free neutron decays to a proton along with an electron and an antineutrino, the

$Q$ value of the reaction is positive which means the reaction is possible all by itself. Hence, statement A is correct.

A free proton cannot convert itself into a neutron due to negative

$Q$ value. Hence, statement B is incorrect.
In beta minus decay, the electron originates from nucleus only, by the transformation of neutron into a proton, with simultaneous emission of an antineutrino, this reaction is same as the first one and so it is feasible. Hence, statement C is correct.

Taking the Bohr radius as

$a_0 = 53 pm$ , the radius of

$Li^{++}$ ion in its ground state, on the basis of Bohr's model, will be about

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$53 pm$
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$27 pm$
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$18 pm$
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$13 pm$

Explanation

According to Bohr's model of atom, radius of an atom in its ground state is

$r = \dfrac{r_o}{z}$ where

$r_o$ is Bohr's radius, and Z is atomic number.

$r_o = 53 pm$ and atomic number of Lithium atom is

$3$ so,

$r = \dfrac{53}{3} = 17.67 pm = 18 pm$ verifies option (c).

Which of the following statements about Rutherford’s model of atom are correct?

(i) Proposed that nearly all the mass of an atom resides in the nucleus.
(ii) Established that the

$\alpha$ –particles are four times as heavy as a hydrogen atom.
(iii) Considered the nucleus as positively charged.
(iv) Agreed with Thomson’s model.

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(i) and (iii)
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(ii) and (iii)
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(i) and (iv)
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only (i)

Rutherford's

$\alpha-$ particle scattering experiment discovered

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Electron
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Proton
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Atomic nucleus
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Neutron

Explanation

Nucleus of the atom that is positively charged was discovered by Rutherford's

$\alpha-$ particle scattering experiment.

In the Bohr's hydrogen atom model, the radius of the stationary orbit is directly proportional to (

$n=$ principle quantum number)

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$n^{-1}$
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$n$
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$n^{-2}$
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$n^{2}$

Explanation

Bohr radius

$r=\dfrac{\varepsilon_{0}n^{2}h^{2}}{\pi Zme^{2}};$

$\therefore r\propto n^{2}$

The Balmer series for the H-atom can be observed

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if we measure the frequencies of light emitted when an excited atom falls to the ground state.
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if we measure the frequencies of light emitted due to transitions between excited states and the first excited state.
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in any transition in a H - atom
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as a sequence of frequencies with the higher frequencies getting closely packed.

According to Bohr's theory the radius of electron in an orbit described by principle quantum number

$n$ and atomic number

$Z$ is proportional to

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$Z^{2}n^{2}$
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$\dfrac{Z^{2}}{n^{2}}$
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$\dfrac{Z^{2}}{n}$
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$\dfrac{n^{2}}{Z}$

Explanation

$r=\dfrac{\varepsilon_{0}n^{2}h^{2}}{\pi Zme^{2}};$

$\therefore r\propto \dfrac{n^{2}}{Z}$

According to de-Broglie wavelength for electron in an orbit of hydrogen atom is

$10^{-9}\ m$ . The principle quantum number for this electron is

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$1$
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$2$
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$3$
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$4$

Explanation

As we know,

$2\pi rn= \lambda$

$\Rightarrow n=\dfrac {\lambda}{2\pi r}=\dfrac {10^{-9}}{2\times 3.14 \times 5.13\times 10^{-11}}=3$

Which of the following is true.

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Lyman series is a continuous spectrum
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Paschen series is a line spectrum in the infrared
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Balmer series is a line spectrum in the ultraviolet
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The spectral series formula can be derived from the Rutherford model of the hydrogen atom

The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the

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Balmer series
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Pfund series
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Paschen series
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Lyman series

The splitting of line into groups under the effect of electric or magnetic field is called

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Zeeman's effect
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Bohr's effect
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Heisenberg's effect
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Magnetic effect

In bohr's model, if the atomic radius of the first orbit is

$r_0$ , then the radius of the fourth orbit is

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$r_0$
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$4r_0$
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$r_0/16$
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$16r_0$

Explanation

$r_n \propto n^2\Rightarrow \dfrac {r_4}{r_1}=\left( \dfrac 41 \right)^2 =\dfrac {16}{1}\Rightarrow r_4=16r_1 \Rightarrow r_4=16\ r_0$

To explain his theory, Bohr-used

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Conservation of linear momentum
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Conservation of angular momentum
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Conservation of quantum frequency
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Conservation of energy

Which one of these is non-divisible

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Nucleous
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Photon
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Proton
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Atom

Radius of the first orbit of the electron in a hydrogen atom is

$0.53\ A^o$ . So, the radius of the third orbit will be

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$2.12\ A^o$
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$4.77\ A^o$
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$1.06\ A^o$
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$1.59\ A^o$

Explanation

$r_n \propto n^2 \Rightarrow \dfrac{r_3}{r_1}=\dfrac{3^2}{1}$

$\Rightarrow r_3=9r_1=9\times 0.53 =4.77\ A^o$

The fact that photons carry energy was established by

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Doppler's effect
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Compton's effect
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Bohr's theory
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Diffraction of light

Which of the following spectral series in hydrogen atom give spectral line of

$4860\ A^o$

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Lyman
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Balmer
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Paschen
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Brackett

For principal quantum number

$n=3$ , the possible values of orbital quantum number

$'I$ are

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$1, 2, 3$
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$0, 1, 2, 3$
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$0, 1, 2$
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$-1, 0, +1$

Explanation

For

$M$ shell

$(n=3)$ , orbital quantum number

$I=0, 1, 2$ .

In Bohr's model of hydrogen atom, which of the following pairs of quantities are quantized

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Energy and linear momentum
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Linear and angular momentum
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Energy and angular momentum
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None of these

The boher model of atoms

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Assume that the angular momentum of electrons is quantized
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Uses Einstein's photo electric equation
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Predicts continuous emission spectra for atoms
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Predicts the same emission spectra for all types of atoms

The possible quantum number for

$3d$ electron are

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$n=3, l=1, m_1=+1, m_s=-\dfrac 12$
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$n=3, l=2, m_1=+2, m_s=-\dfrac 12$
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$n=3, l=1, m_1=-1, m_s=+\dfrac 12$
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$n=3, l=0, m_1=+1, m_s=-\dfrac 12$

The radius of the first ( lowest ) orbit of the hydrogen atom is

$a_0$ . The radius of the second ( next higher ) orbit will be

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$4a_0$
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$6a_0$
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$8a_0$
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$10a_0$

Explanation

$r_n=\dfrac{n^2}{z}a_b$

$r_1=\dfrac{1}{L}a_0$

$r_1=a_0$

$r_2=\dfrac{(2)^2}{1}a_0$

$r_2=4a_0$

Radius of first Bohr orbit is

$r$ . What is the radius of

$2-$ Bohr orbit?

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$8\ r$
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$2\ r$
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$4\ r$
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$2\ \sqrt {2r}$

In hydrogen atom which quantity is integral multiple of

$\dfrac{h}{2\pi}$

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Angular momentum
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Angular velocity
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Angular acceleration
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Momentum

The diagram shows the path of four

$\alpha$ - particles of the same energy being scattered by the nucleus of an atom simultaneously. Which of these are/ is not physically possible

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$3$ and $4$
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$2$ and $3$
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$1$ and $4$
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$4$ only

Explanation

$\alpha-$ particles cannot be attracted by the nucleus.

Hence, option 4 is correct option.

Line spectrum contains information about

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The atoms of the prism
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The atoms of the source
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The molecules of the source
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The atoms as well as molecules of the source

Line spectra are due to

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Hot solids
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Atoms in gaseous state
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Molecules in gaseous state
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Liquid at low temperature

The spectrum obtained from a sodium vapour lamp is an example of

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Absorption spectrum
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Emission spectrum
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Continuous spectrum
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Band spectrum

The solar spectrum during a complete solar eclipse is

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Continuous
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Emission line
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Dark line
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Dark band

In which part of the electromagnetic spectrum the Lyman series of hydrogen is found ?

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ultraviolet
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infrared
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visible
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X-ray region

The nature of suns spectrum is

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Continuous spectrum with absorption lines
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Line spectrum
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The spectrum of the helium atom
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Band spectrum

When a hydrogen atom is raised from ground to excited state.

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The P.E increases and K.E decreases
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The P.E decreases and K.E increases
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Both P.E and K.E increase
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Both P.E and K.E decreases

Explanation

Potential energy of hydrogen atom is given by

$P.E=\dfrac{-kZe}{r}$

As atom is excited,it's radius increases. But due to negative sign or reverse sign. Its actual potential energy increases.

And kinetic energy is given by

$K.E=\dfrac{kZe}{2r}$

As Tom is excited, it's radius increases this resulting in decrease of kinetic energy.

The PE increases and KE decreases.

In which of the following systems will the radius of the first orbit of the electron be smallest?

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hydrogen
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singly ionized helium
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deuteron
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tritium

In which of these, will radius of first orbit be minimum?

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$H$ atom
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$^{2}_{1}H$ atom
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$He^{+}$ ion
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$Li^{2+}$ ion

Explanation

$r_{n}^{1}=4\pi \epsilon _{0}\dfrac{n^{2}h^{2}}{4\pi ^{2}z.me^{2}}$

$Z$ is max for

$Li^{2+}$ ion.

$\therefore$ radius is minimum for

$Li^{2+}$ ion.

The allowed momenta are given by

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$\dfrac{nh}{2L}$
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$\dfrac{nh}{L}$
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$\dfrac{nh}{2\Pi L}$
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$\dfrac{nh}{4L}$

Explanation

From Bohr-Sommerfeld Quantization condition given in the paragraph, for a back and forth motion.

$P\times 2L=nh$ , where P is the momentum,

$P=\dfrac{nh}{2L}$

The allowed kinetic energy of the particle is

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$\dfrac{n^{2}h^{2}}{8\pi ^{2}mL^{2}}$
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$\dfrac{n^{2}h^{2}}{2mL^{2}}$
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$\dfrac{n^{2}h^{2}}{8mL^{2}}$
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$\dfrac{n^{2}h^{2}}{32mL^{2}}$

Explanation

From Bohr-Sommerfeld Quantization condition given in the paragraph, for a back and forth motion.

$P\times 2L=nh$ , where P is the momentum,

$P=\dfrac{nh}{2L}$

And the kinetic energy,

$E=\dfrac{P^{2}}{2m}$

$E=\dfrac{n^{2}h^{2}}{8mL^{2}}$

Assume an imaginary world where angular momentum is quantized to even multiple. The longest possible wave length emitted by hydrogen in the visible spectrum is

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$484 nm$
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$300 nm$
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$400 nm$
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$350 nm$

Explanation

The angular momentum is quantized to even multiple of

$\bar h$ , hence

$mvr = 2n \hbar$

$mv = \dfrac{2n\hbar }{r}$

$mv^2 = \dfrac{m^2v^2}{m} = \dfrac{(2n\hbar )^2}{mr}$

The Coulomb's force of attraction is equal to the centripetal force.

$\dfrac{Ze^2}{4\pi \epsilon_0 r^2} = \dfrac{mv^2}{r}$

$\dfrac{Ze^2}{4\pi \epsilon_0 r^2} = \dfrac{\dfrac{(2n\hbar )^2}{mr}}{r}$

$r = \dfrac{(2n\hbar )^2 4\pi \epsilon_0}{mZe^2}$

Now, the binding energy of electron is

$BE = \dfrac{-Ze^2}{8\pi \epsilon_0 r}$

$BE = \dfrac{-Ze^2}{8\pi \epsilon_0 (\dfrac{(2n\hbar )^2 4\pi \epsilon_0}{mZe^2})}$

$BE = \dfrac{-Z^2e^4m}{32 \pi \epsilon_0^2n^2h^2}$

$BE = \dfrac{-3.4}{n^2} eV$

For longest possible wavelength in visible spectrum

$n_2 = 4$ and

$n_1 = 1$

$h\nu = -3.4[1 - \dfrac{1}{4}]$

$h\nu = -3.4 (\dfrac{3}{4})$

$\nu = -\dfrac{3.4 \times 0.75}{h}$

Therefore, the longest possible wavelength emitted by hydrogen in the visible spectrum is(in nm)

$\lambda = \dfrac{c}{\nu}$

$\lambda = \dfrac{hc}{3.4 \times 0.75}$

$\lambda = \dfrac{1236}{2.55}$

$\lambda = 484\ nm$

A particle of mass

$m$ moves around in a circular orbit in a centro symmetric potential field u(r)

$=\dfrac{kr^{2}}{2}$ . Using Bohr’s quantization rule, the permissible energy levels are

0%
$\sqrt{\dfrac{k}{m}}$
0%
$n\sqrt{\dfrac{k}{m}}$
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$n$
0%
$\sqrt{\dfrac{nk}{m}}$

Explanation

$U=\dfrac{kr^2}{2}$

$\implies F=-\dfrac{mv^2}{r}=\dfrac{-dU}{dr}=-kr$

$\implies \dfrac{1}{2}mv^2=\dfrac{1}{2}kr^2$

Total energy

$=\dfrac{1}{2}mv^2+U=kr^2$

From bohr's quantization,

$mvr=\dfrac{nh}{2\pi}$

$\implies m^2v^2r^2=\dfrac{n^2h^2}{4\pi^2}$

$\implies mkr^4=\dfrac{n^2h^2}{4\pi^2}$

$\implies r^2=\dfrac{nh}{2\pi \sqrt{mk}}$

Energy

$=kr^2=k\dfrac{nh}{2\pi \sqrt{mk}}\propto n\sqrt{\dfrac{k}{m}}$

Let

$\mathrm{v}_{1}$ be the frequency of the series limit for Lyman series,

$\mathrm{v}_{2}$ the frequency of the first line of the Lyman series and

$\mathrm{v}_{3}$ the frequency of the series limit of Balmer series. Then which of the following relations is true?

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$\mathrm{v}_{1}-\mathrm{v}_{2}=\mathrm{v}_{3}$
0%
$\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}_{3}$
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$\displaystyle \mathrm{v}_{3}=\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}$
0%
$\mathrm{v}_{1}+\mathrm{v}_{2}=\mathrm{v}_{3}$

Explanation

Series limit means the shortest possible wavelength (maximum photon energy) and first line means the longest possible wavelength, (minimum photon energy) in the series

$\displaystyle \mathrm{v}=\mathrm{c}[\frac{1}{\mathrm{n}^{2}}-\frac{1}{\mathrm{m}^{2}}]$ , where c is a constant.

For series limit of Lyman series

$\mathrm{n}=1 \mathrm{m}=\infty \mathrm{v}_{1}=\mathrm{c}$

For first line of Lyman series

$\mathrm{n}=1 \mathrm{m}=2 \displaystyle \mathrm{v}_{2}=\frac{3\mathrm{c}}{4}$

For series limit of Balmer serles

$\mathrm{n}=2 \mathrm{m}=\infty \displaystyle \mathrm{v}_{3}=\frac{\mathrm{c}}{4}$
:.

$\mathrm{v}_{1}-\mathrm{v}_{2}=\mathrm{v}_{3}$

$E_1 = E_2 + E_3$

$hV_1 = hV_2 + hV_3$

$V_1 = V_2 + V_3$

The radius of hydrogen atom is 0.53

$\mathop A\limits^0$ . The radius of

$Li^{2+}$ is of

0%
$1.27\mathop A\limits^0$
0%
$0.17 \mathop A\limits^0$
0%
$0.57 \mathop A\limits^0$
0%
$0.99 \mathop A\limits^0$

Explanation

$\begin{array}{l}\text { radius, } r=\frac{n^{2} h^{2}}{4 \pi^{2} m z e^{2}}=0.53\frac{n^{2}}{z} \dot A \\\text { for ground state } n=1 . \\\text { radius hydrogen in ground state } \\\text { = }\frac{0.53(1)^{2}}{1}=0.53 \dot A \\\text { [atomic no. of } \\\text { hydrogen, } z=1]\end{array}$

$\begin{aligned}\text { that of } & L i^{2+}\text { is } \\r=& \frac{0.53 \times(1)^{2}}{3} \\&=0.177 \mathrm{~A} \\& \approx 0.17\mathrm{~A} .\end{aligned}$

Whenever a hydrogen atom emits a photon in the Balmer series,

0%
it may emit another photon in the Balmer series.
0%
it must emit another photon in the Lyman series.
0%
the second photon, it emitted, will have a wavelength of about 122 nm.
0%
it may emit a second photon, but the wavelength of this photon cannot be predicted.

Explanation

For Balmer series,

$n_1 = 2$ and

$n_2 = 3,4,............,\infty$ . Hence, all the transitons takes place from higher energy levels to second level. But this level is unstable hence to achieve stability electrons undergo another transition from second level to first one which is the stable level. Hence, another photon emitted in Lyman series.

The wavelength of photon is given by:

$\dfrac{1}{\lambda} = R[\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}]$

$\dfrac{1}{\lambda} = R[\dfrac{1}{1^2} - \dfrac{1}{2^2}]$

$\dfrac{1}{\lambda} = R[1 - \dfrac{1}{4}]$

$\dfrac{1}{\lambda} = R[\dfrac{3}{4}]$

$\lambda = \dfrac{4}{3R}$

$\lambda = \dfrac{4}{3 \times 10.96 \times 10^{6}}$

$\lambda = \dfrac{4}{32.88 \times 10^{6}}$

$\lambda = 0.1217 \times 10^{-6}$

$\lambda = 121.7 \approx 122\ nm$

When

$Z$ is doubled in a hydrogen like atom, which of the following statements are consistent with Bohr's theory?

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Energy of a state is double
0%
Radius of an orbit is doubled
0%
Velocity of electrons in an orbit is doubled
0%
Radius of an orbit is halved

Explanation

From the concepts attached,

$E \propto z^2$

$V \propto Z$

$r \propto \cfrac{1}{Z}$

$T \propto \cfrac{1}{Z^2}$
hence,
Velocity doubles and radius halved

The difference between I and II energy levels is

0%
$1.5\times 10^{-18} J$
0%
$3\times 10^{-18} J$
0%
$6\times 10^{-18} J$
0%
$12\times 10^{-18} J$

Explanation

For this particle in the box:

$E=\dfrac{n^{2}h^{2}}{8mL^{2}}$

The difference between I and II energy levels is:

$E_{diff}=\dfrac{3h^{2}}{8mL^{2}}=1.5 \times 10^{-18} J$

CBSE Questions for Class 12 Medical Physics Atoms Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Atoms Quiz 12 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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