Physics - Atoms - Quiz-2

In terms of Bohr radius $a_{0}$ , the radius of the second Bohr orbit of a hydrogen atom is given by [1992]

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4 $a_{0}$
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8 $a_{0}$
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$\sqrt{2}$ $a_{0}$
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2 $a_{0}$

An x-ray tube is operating at 30 kV then the minimum wavelength of the x-rays coming out of the tube is:

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1.24 $\overset{0}{A}$
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0.413 $\overset{0}{A}$
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0.124 $\overset{0}{A}$
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0.13 $\overset{0}{A}$

A diatomic molecule is made of two masses $m_{1} a n d m_{2}$ which are separated by a distance r. If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by (n is an integer):

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$\frac{{(m_{1} + m_{2})}^{2} n^{2} h^{2}}{2 m_{1}^{2} m_{2}^{2} r^{2}}$
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$\frac{n^{2} h^{2}}{2 π^{2} (m_{1}^{} + m_{2}^{}) r^{2}}$
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$\frac{2 n^{2} h^{2}}{(m_{1}^{} + m_{2}^{}) r^{2}}$
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$\frac{(m_{1} + m_{2}) n^{2} h^{2}}{8 π^{2} m_{1}^{} m_{2}^{} r^{2}}$

In a hydrogen atom, which of the following electronic transitions would involve the maximum energy change ?

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From n = 3 to n = 1
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From n = 4 to n = 2
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From n = 3 to n = 2
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From n = 2 to n = 1

The Bohr model of the atom:

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Assumes that the angular momentum of electrons is quantized
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Uses Einstein's photoelectric 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

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

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both KE and PE increase
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both KE and PE decrease
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PE increases and KE decreases
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PE decreases and KE increases

The extreme wavelengths of Paschen series are

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0.365 $μ m$ and 0.565 $μ m$
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0.818 $μ m$ and 0.189 $μ m$
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1.45 $μ m$ and 4.04 $μ m$
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2.27 $μ m$ and 7.43 $μ m$

A hydrogen atom is in an excited state of principal quantum number (n). It emits a photon of wavelength ( $λ$ ) when it returns to the ground state. The value of n is:

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$\sqrt{\frac{λ R}{λ R - 1}}$
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$\sqrt{\frac{(λ R - 1)}{λ R}}$
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$\sqrt{λ (R - 1)}$
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None of these

Consider an electron in the nth orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of the de-Broglie wavelength $λ$ of that electron as:

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(0.529) n $λ$
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$\sqrt{n λ}$
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(13.6) $λ$
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n $λ$

In the Bohr's model of a hydrogen atom, the centripetal force is furnished by the Coulomb attraction between the proton and the electron. If $α_{0}$ is the radius of the ground state orbit, m is the mass and e is the charge on the electron, $ε_{0}$ is the vaccum permittivity, the speed of the electron is [1998]

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zero
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$\frac{e}{\sqrt{ε_{0} a_{0} m}}$
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$\frac{e}{\sqrt{4 {πε}_{0} a_{0} m}}$
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$\frac{\sqrt{4 {πε}_{0} a_{0} m}}{e}$

The ground state energy of H-atom is 13.6 eV. The energy needed to ionise H-atom from its second excited state [1991]

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1.51 eV
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3.4 eV
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13.6 eV
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12.1 eV

The given diagram indicates the energy levels of a certain atom. When the system moves from 2E level to E, a photon of wavelength $λ$ is emitted. The wavelength of photon produced during its transition from $\frac{4 E}{3}$ level to E is

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$\frac{λ}{3}$
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$\frac{3 λ}{4}$
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$\frac{4 λ}{3}$
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$3 λ$

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 $\overset{0}{A}$ . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is [IIT-JEE 2011]

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1215 $\overset{0}{A}$
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1640 $\overset{0}{A}$
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2430 $\overset{0}{A}$
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4687 $\overset{0}{A}$

Energy E of a hydrogen atom with principal quantum number n is given by $E = \frac{- 13.6}{n^{2}} e V$ . The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen, is approximately [2004]

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1.5 eV
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0.85 eV
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3.4 eV
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1.9 eV

The energy of ground electronic state of hydrogen atom is -13.6 eV. The energy of the first excited state will be [1997]

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-54.4 eV
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-27.2 eV
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-6.8 eV
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-3.4 eV

The total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its kinetic energy in this state is [2005]

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-3.4 eV
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-6.8 eV
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6.8 eV
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3.4 eV

When hydrogen atom is in its first excited level, its radius is [1997]

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four times, its ground state radius
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twice, its ground state radius
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same as its ground state radius
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half of its ground state radius

The ionisation potential of helium atom is 24.6 volt, the energy required to ionise it will be

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24.6 eV
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24.6 volt
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13.6 volt
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13..6 eV

The ionization energy of 10 times ionized sodium atom is:

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13.6 eV
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$13.6 \times 11 e V$
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$\frac{13.6}{11} e V$
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$13.6 \times {(11)}^{2} e V$

The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be [1988]

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13.6 eV
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27.2 eV
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6.8 eV
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54.4 eV

The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen-like atom X in the second excited state. As a result the hydrogen-like atom X makes a transition to $n^{t h}$ orbit. Then:

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X = $H e^{+}, n = 4$
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X = $L i^{+ +}$ , n = 6
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X = $H e^{+}, n = 9$
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X = $L i^{+ +}$ , n = 9

If an electron in an hydrogen atom jumps from an orbit $n_{i} = 3$ to an orbit with level $n_{f} = 2$ , the frequency of the emitted radiation is

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$v = \frac{36 C}{5 R}$
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v = $\frac{C R}{6}$
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v = $\frac{5 R C}{36}$
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v = $\frac{6 C}{R}$

The wavenumber of a photon in the Brackett series of a hydrogen atom is $\frac{9}{400} R$ . The electron has transited from the orbit having quantum number: [Odisha JEE 2012]

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5
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6
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4
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7

In an experiment to determine the e/m value for an electron using Thomson's method the electrostatic deflection plates were 0.01 m apart and had a potential difference of 200 volts applied. Then the electric field strength between the plates is

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$1 \times 10^{4} V / m$
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$2 \times 10^{4} V / m$
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$4 \times 10^{4} V / m$
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$5 \times 10^{5} V / m$

Hydrogen $({}_{1}H_{1})$ , Deuterium $({}_{1}H_{2})$ , singly ionised Helium ${({}_{2}H e^{4})}^{+}$ , and doubly ionised lithium ${({}_{3}L i^{6})}^{+ +}$ all have one electron around the nucleus. Consider an electron transition from n = 2 to n = 1. If the wavelengths of emitted radiations are $λ_{1}, λ_{2}, λ_{3}$ and $λ_{4}$ respectively, then approximately which one of the following is correct?
[JEE (Main) 2014]

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$4 λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$
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$λ_{1} = 2 λ_{2} = 2 λ_{3} = λ_{4}$
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$λ_{1} = λ_{2} = 4 λ_{3} = 9 λ_{4}$
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$λ_{1} = 2 λ_{2} = 3 λ_{3} = λ_{4}$

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

Bragg's law for X-rays is:

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dsin $θ$ = 2n $λ$
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2dsin $θ$ = n $λ$
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nsin $θ$ = 2 $λ$ d
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None of these

The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volt is equal to

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$\frac{e V}{h c}$
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$\frac{e h}{c V}$
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$\frac{h c}{e V}$
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$\frac{c V}{e h}$

The graph between the square root of the frequency of a specific line of the characteristic spectrum of X-rays and the atomic number of the target will be:

The wavelength of $K_{α}$ X-rays for lead isotopes $P b^{208}, P b^{206}, P b^{204} a r e λ_{1}, λ_{2} a n d λ_{3}$ respectively. Then

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$λ_{1} = λ_{2} > λ_{3}$
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$λ_{1} > λ_{2} > λ_{3}$
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$λ_{1} < λ_{2} < λ_{3}$
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$λ_{2} = λ_{1} = λ_{3}$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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