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QUESTION: 1

What is the radius of iodine atom (At. no. 53,mass no. 126) [1988]

Solution:

53 electrons in iodine atom are distributed as

2, 8, 18, 18, 7

∴ n = 5

QUESTION: 2

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

Solution:

E ∝ Z^{2} and Z for helium = 2

∴ (E)_{He} = 4x13.6 = 54.4 eV

QUESTION: 3

To explain his theory, Bohr used [1989]

Solution:

Bohr used conservation of angular

momentum.

QUESTION: 4

The ground state energy of H-atom 13.6 eV. Theenergy needed to ionize H-atom from its secondexcited state. [1991]

Solution:

Second excited state corresponds to n = 3

QUESTION: 5

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

Solution:

As r ∝ n^{2} , therefore, radius of 2nd

Bohr’s orbit = 4 a_{0}.

QUESTION: 6

The ionization energy of hydrogen atom is 13.6eV. Following Bohr’s theory, the energycorresponding to a transition between 3rd and4th orbit is [1992]

Solution:

QUESTION: 7

Which source is associated with a line emissionspectrum ? [1993]

Solution:

Neon street sign is a source of line emission

spectrum.

QUESTION: 8

In Rutherford scattering experiment, what willbe the correct angle for α-scattering for an impactparameter, b = 0 ? [1994]

Solution:

Impact parameter for Rutherford scattering

experiment,

QUESTION: 9

The radius of hydrogen atom in its ground stateis 5.3 × 10^{–11} m. After collision with an electron itis found to have a radius of 21.2 × 10^{–11} m. Whatis the principal quantum number n of the finalstate of the atom [1994]

Solution:

r ∝ n^{2}

QUESTION: 10

The spectrum obtained from a sodium vapourlamp is an example of [1995]

Solution:

A spectrum is observed, when light coming

directly from a source is examined with a

spectroscope. Therefore spectrum obtained

from a sodium vapour lamp is emission

spectrum.

QUESTION: 11

When a hydrogen atom is raised from the groundstate to an excited state, [1995]

Solution:

and where,

r is the radius of orbit which increases as we

move from ground to an excited state.

Therefore, when a hydrogen atom is raised

from the ground state, it increases the value

of r. As a result of this, P.E. increases

(decreases in negative) and K.E. decreases.

QUESTION: 12

If the threshold wavelength for a certain metal is 2000 Å, then the work-function of the metal is [1995]

Solution:

Threshold wavelength (λ) = 2000 Å = 2000 ×

10^{–10} m. Work function

QUESTION: 13

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

Solution:

Radius in ground state =

Radius in first excited state =

(∵ n =2)

Hence, radius of first excited state is four

times the radius in ground state.

QUESTION: 14

In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If a_{0} is the radius of the ground state orbit, m is the mass, e is the charge on the electron and ε_{0} is the vacuum ermittivity, the speed of the electron is [1998]

Solution:

Centripetal force = Coulombian force

QUESTION: 15

Who indirectly determined the mass of theelectron by measuring the charge of the electron? [2000]

Solution:

**Correct Answer :- D**

**Explanation : **The mass of the electron is discovered by the Millikan by the oil drop experiment.

E = mg

=> qE = mg

q/m = g/E

Therefore, charge was quantised.

QUESTION: 16

When an electron jumps from the fourth orbit tothe second orbit, one gets the [2000]

Solution:

When the electron drops from any orbit to

second orbit, then wavelength of line

obtained belongs to Balmer series.

QUESTION: 17

Which of the following transitions in a hydrogenatom emits the photon of highest frequency? [2000]

Solution:

Frequency,

Note : See the greatest energy difference and

also see that the transition is from higher to

lower energy level. Hence, it is highest in

case of n = 2 to n = 1.

QUESTION: 18

An electron changes its position from orbit n = 2 to the orbit n = 4 of an atom. The wavelength of the emitted radiations is (R = Rydberg’s constant)

Solution:

, where n_{1} = 2, n_{2} = 4

QUESTION: 19

The energy of hydrogen atom in nth orbit is En, then the energy in nth orbit of single ionised helium atom will be [2001]

Solution:

We have For helium

Z = 2. Hence requisite answer is 4E_{n}.

QUESTION: 20

J.J. Thomson’s cathode-ray tube experimentdemonstrated that [2003]

Solution:

Cathode rays are streams of negatively

charged ions

QUESTION: 21

In which of the following systems will the radiusof the first orbit (n = 1) be minimum ? [2003]

Solution:

Z(=3) is maximum for Li^{2+}.

QUESTION: 22

The Bohr model of atoms [2004]

Solution:

In Bohr’s model, angular momentum is

quantised i.e =

QUESTION: 23

The Bohr model of atomsEnergy E of a hydrogen atom with principalquantum number n is given by E = – 13.6/n^{2} eV.The energy of photon ejected when the electronjumps from n = 3 state to n = 2 state of hydrogenis approximately [2004]

Solution:

QUESTION: 24

The half life of radium is about 1600 years. Of 100 g of radium existing now, 25 g will remainunchanged after [2004]

Solution:

The total time in which radium change to 25 g is

= 2 x 1600

= 3200 yr.

QUESTION: 25

The total energy of an electron in the firstexcited state of hydrogen atom is about –3.4eV. Its kinetic energy in this state is [2005]

Solution:

Mechanical energy =

∴ K.E. in 2nd orbital for hydrogen

= – Mechanical energy

QUESTION: 26

Ionization potential of hydrogen atom is 13.6eV.Hydrogen atoms in the ground state are excitedby monochromatic radiation of photon energy12.1 eV. According to Bohr’s theory, the spectrallines emitted by hydrogen will be [2006]

Solution:

Energy of ground state 13.6 eV

Energy of first excited state

Energy of second excited state

Difference between ground state and 2nd

excited state = 13.6 – 1.5 = 12.1 eV

So, electron can be excited upto 3rd orbit

No. of possible transition

1 → 2, 1 → 3, 2 → 3

So, three lines are possible

QUESTION: 27

The total energy of electron in the ground stateof hydrogen atom is – 13.6 eV. The kinetic energyof an electron in the first excited state is [2007]

Solution:

Energy in the first excited state

But K.E. = –(Total energy) = +3.4 eV.

QUESTION: 28

The ground state energy of hydrogen atom is 13.6eV. When its electron is in the first excited state, its excitation energy is [2008]

Solution:

When the electron is in first excited state

(n = 2), the excitation energy is given by

ΔE = E_{2} – E_{1}

Given E1 = –13.6eV

∴Δ E = ( – 3.4) – ( – 13.6) = 10.2 eV.

QUESTION: 29

In a Rutherford scattering experiment when a projectile of charge Z_{1} and mass M_{1} approaches a target nucleus of charge Z_{2} and mass M_{2}, the distance of closest approach is r_{0}. The energy of the projectile is [2009]

Solution:

The kinetic energy of the projectile is given by

Thus energy of the projectile is directly

proportional to Z_{1}, Z_{2}

QUESTION: 30

The ionization energy of the electron in thehydrogen atom in its ground state is 13.6 eV.The atoms are excited to higher energy levels toemit radiations of 6 wavelengths. Maximumwavelength of emitted radiation corresponds tothe transition between [2009]

Solution:

n^{2} – n – 12 = 0

(n – 4) (n + 3) = 0 or n = 4

QUESTION: 31

The energy of a hydrogen atom in the groundstate is – 13.6 eV. The energy of a He+ ion in thefirst excited state will be [2010]

Solution:

Energy of a H-like atom in it's nth state is

given by

For, first excited state of He^{+}, n = 2, Z = 2

QUESTION: 32

An alpha nucleus of energy 1/2mv^{2 }bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to [2010]

Solution:

Kinetic energy of alpha nucleus is equall to

electrostatic potential energy of the system

of the alpha particle and the heavy nucleus.

That is,

where r_{0} is the distance of closest approach

Hence, correct option is (c).

QUESTION: 33

The wavelength of the first line of Lyman seriesfor hydrogen atom is equal to that of the secondline of Balmer series for a hydrogen like ion. Theatomic number Z of hydrogen like ion is [2011]

Solution:

For first line of Lyman series of hydrogen

For second line of Balmer series of hydrogen

like ion

QUESTION: 34

An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a

photosensitive material having work function 2.75 eV. If the stopping potential of thephotoelectron is 10 V, the value of n is [2011 M]

Solution:

QUESTION: 35

Out of the following which one is not a possibleenergy for a photon to be emitted by hydrogenatom according to Bohr’s atomic model? [2011M]

Solution:

Obviously, difference of 11.1eV is not

possible.

QUESTION: 36

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state.The ratio of the wavelength λ_{1} : λ_{2} emitted inthe two cases is [2012]

Solution:

QUESTION: 37

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be : [2012]

(m is the mass of the electron, R, Rydberg constant and h Planck’s constant)

Solution:

For emission, the wave number of the

radiation is given as

R= Rydberg constant, Z = atomic number

linear momentum

(de-Broglie hypothesis)

QUESTION: 38

The transition from the state n = 3 to n = 1 in ahydrogen like atom results in ultravioletradiation. Infrared radiation will be obtained inthe transition from : [2012M]

Solution:

The frequency of the transition

when n = 1, 2, 3.

QUESTION: 39

The ratio of longest wavelengths corresponding to Lyman and Blamer series in hydrogen spectrum is [NEET 2013]

Solution:

For Lyman series (2 → 1)

For Balmer series (3 → 2)

QUESTION: 40

An electron in hydrogen atom makes a transition n_{1} → n_{2} where n_{1} and n_{2} are principal quantum numbers of the two states. Assuming Bohr’s model to be valid the time period of the electron in the initial state is eight times that in the final state. The possible values of n_{1} and n_{2} are [NEET Kar. 2013]

Solution:

T ∝ n^{3}

Tn1 = 8 Tn_{2} (given)

Hence, n_{1} = 2n_{2}

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