Atoms

The difference in the angular momentum associated with the electron in the 3rd and 4th orbits of the hydrogen atom is ____.

Correct! Wrong!

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According to Bohr's postulate, the angular momentum of a orbiting electron in the nth orbit is, Ln=nh/2π
Angular momentum of the electron in the 3rd orbit is, L3=3h/2π
Angular momentum of the electron in the 4th orbit is, L4=4h/2π
Then the difference in the angular momentum is L4−L3= (4h/2π)−(3h/2π)
=h/2π

The colour of the second line of the Balmer series is ____.

Correct! Wrong!

The empirical formula for Balmer series is, 1/λ=R(1/22−1/n2)
The second line of the Balmer series is obtained for n = 4
we know that, R=1.097×107m−1
Then 1/λ=1.097×107(1/22−1/42)
1/λ=3/16×1.097×107
λ=4.8617×10−7m
= 4861.7 A
This wavelength represents blue colour of the visible light.

The energy of electron in the first orbit of hydrogen like atom of atomic number, Z is proportional to ____.

Correct! Wrong!

The energy of electron in the nth orbit of hydrogen like atom of atomic number Z is, En= (−13.6/n2)Z2 eV

An electron makes a transition from 4th orbit to the 2nd orbit of hydrogen atom. The wave number of the emitted radiation is ____. ( R is the Rydberg's constant)

Correct! Wrong!

Wave number, 1/λ=R(1/n12−1/n22)
Given, n1=2 and n2=4
Then, 1/λ=R(1/22−1/42)
1/λ=R×3/16
Thus the wave number is 3R/16.

If the energy of electron in the first orbit of hydrogen atom is E, the energy of the electron in the 2nd orbit of He+ atom is ____

Correct! Wrong!

The energy of electron in the nth orbit of hydrogen like atom of atomic number Z is, En=−(13.6/n2 )Z2eV
The energy of electron in the first orbit of hydrogen atom
Is, E=− (13.6/12 )12eV
E=−13.6eV
The energy of electron in the 2nd orbit of He+ atom is, E2=− (13.6/22 ) ×22eV
E2=−13.6eV
Thus E2=E

The ratio of size of an atom to size of a nucleus is ____.

Correct! Wrong!

We know that,
Size of an atom is 10−10m
Size of a nucleus is 10−14m
Then the ratio of size of an atom to the size of a nucleus is, 10−10/10−14
=104:1

If Kinetic Energy of electron in hydrogen atom is K and the Total Energy of the electron is E. the relation between K and E is ____.

Correct! Wrong!

We know that,
The kinetic energy of electron in hydrogen atom, K =e2/8πϵ0r
The total energy of the electron in hydrogen atom, E =−e2/8πϵ0r
On comparing the above two equations, E=−K

Which of the following two elements do not have identical line spectrum?

Correct! Wrong!

Each element has a characteristic spectrum. No two elements can give the same line spectrum. Each element has its own line spectrum. Line spectrum is also called finger print spectrum.

In alpha particle scattering, if the kinetic energy of alpha particle is 9 MeV. The distance of closest approach of the nucleus is ____.

Correct! Wrong!

Given the kinetic energy of the alpha particle, K.E = 9 MeV
K.E = 9 × 1.6 × 10−13J
At the distance of closest approach, the alpha particle comes to rest and its kinetic energy is equal to the potential energy of the system.
Potential Energy of the system, P.E = 1/4πϵ0×[(2e)(Ze])/d
where Z is the atomic number of gold and d is the distance of closest approach.
But, K.E = P.E
9 × 1.6 ×10−13 = (1/4πϵ0)×[(2e)(Ze)]/d
we know that, 1/4πϵ0=9×109Nm2C−2
Then, d = 9×109×(2×79×1.6×10−19×1.6×10−19/9×1.6×10−13)
d = 252.8 × 10−16m
d = 25.28 × 10−15m
d= 25.28 fm

The longest wavelength of Lyman series for hydrogen is ____.

Correct! Wrong!

The empirical formula for Lyman series is
1/λ=R(1/12−1/n2)
The longest wavelength of the Lyman series is obtained when the n =2.
We know that the Rydberg constant, R = 1.097 × 107m−1
Then, 1/λ=1.097×107(1/12−1/22)
= 1.097 × 107(1−1/4)
= 0.82275 × 107m−1
Then λ = 1.215 × 10−7m
= 1215 A

Plum pudding model of the atom was proposed by ____.

Correct! Wrong!

The first atomic model was proposed by J. J. Thomson. According to this model, the positive charge of an atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. Thus the model of the atom is known as Plum pudding model of the atom.

The wavelength of the first line of Lyman series of hydrogen is λ. The wavelength of the first line of the Balmer series is ____.

Correct! Wrong!

The empirical formula of Lyman series is, 1/λ=R (1/12−1/n2)
The first line of the Lyman series is obtained for n = 2.
Then 1/λ=3R/4
λ=4/3R
The empirical formula of Balmer series is, 1/λ=R(1/22−1/n2)
The first line of the Balmer series is obtained for n = 3.
Then 1/λ1=R (1/22−1/32)
1/λ1=5R/36
λ1=36/5R
λ1/λ=(36/5R)×(3R/4)
λ1/λ =27/5
λ1=(27/5)λ

According to Bohr's atomic model, the angular momentum of an electron in the nth orbit is ____.

Correct! Wrong!

According to Bohr's second postulate, the electron revolves round the nucleus only in those orbits for which the angular momentum is integral multiple of h/2π.Thus the angular momentum of the orbiting electron is quantised.

The radius of the first orbit of hydrogen atom is called Bohr's radius, a0. The radius of the first orbit of Li+ is___.

Correct! Wrong!

We know that the Bohr's radius, a0=0.529A
For hydrogen like atoms, the radius of nth orbit is, r=(0.529/Z)n2 A
For Li+ atom, Z=3
For first orbit, n=1
Thus the radius of the first orbit of Li+ atom is, r=(0.529/3)×12
=0.529/3 A
=a0/3

The ratio of radii of the first two orbits of hydrogen atom is____.

Correct! Wrong!

We know that the Bohr's radius = (0.529×10−10) ×n2m
For the first orbit, n = 1.
Then the Bohr's radius is, r1= (0.529×10−10) ×12m
For the second orbit, n = 2.
Then the Bohr's radius is, r2= (0.529×10−10) ×22m
Then the ratio of radii of the first orbit to the second orbit is r1/r2=1/4
r1: r2=1:4

The frequency of the radiation emitted by the hydrogen in the transition from first excited state to the ground state is ____.

Correct! Wrong!

For the ground state, n=1
For the first excited state, n=2
We know that, the energy of electron of hydrogen atom in the nth orbit is, En=−13.6/n2eV
The energy of the electron in the ground state is, E1=−13.6/12
E1=−13.6eV
The energy of the electron in the first excited state is, E2=−13.6/22
E2=−3.4eV
According Bohr's atomic model, during the electron transition, the difference in the energy is released or absorbed in the form of radiation.
E2 −E1 = hν
E2 −E1 = −3.4 - (−13.6)
E2 − E1 = 10.2 eV
hν=10.2eV
=10.2×1.6×10−19J
Then the frequency of the radiation, ν=10.2×1.6×10−19/h
=10.2×1.6×10−19 /6.6×10−34Js
=2.4727×1015Hz

According to classical theory, the Rutherford's model of the atom was ____.

Correct! Wrong!

The Rutherford's model is unstable according to classical theory

The ratio of longest wavelength to shortest wavelength of the Balmer series for hydrogen is ____.

Correct! Wrong!

The empirical formula for Balmer series is, 1/λ=R (1/22−1/n2)
The longest wavelength of the series is obtained for n = 3,
1/λ1=R (1/22−1/32)
1/λ1=5R/36
The shortest wavelength of the series is obtained for n = ∞,
1/λ2=R (1/22−1/∞2)
1/λ2=R/4
Then the ratio of longest wavelength to the shortest wavelength of the series is, λ12= (36/5R)×(R/4)
λ12=9/5

If Rutherford's scattering experiment is conducted with different projectiles of charges q1 and q2 moving with the same velocity and the masses of the projectiles are M1 and M2 for the same target. The ratio of the distance of closest approach of the nucleus is ____.

Correct! Wrong!

Given,
The charges of the projectiles are q1 and q2
The masses of the projectile are M1 and M2
At the distance of closest approach,
Kinetic Energy of the projectile = Potential Energy of the system
For the first projectile,
1/2M1v2=1/4πϵ0×[(q1e)(ze)]/d1 .... (1)
where Z is the atomic number of the target and d1is the distance of the closest approach with first projectile
For the second projectile,
1/2M2v2=1/4πϵ0× [(q2e)(Ze)]/d2 ...(2)
Divide the equations (1) and (2).
Then, M1/M2=q1/q2×d2/d1
M1 /M2×q2 /q1=d2/d1
Thus d1/d2=M2q1/M1q2

An electron jumps from3rd orbit to the 2ndorbit of hydrogen atom. The frequency of the emitted radiation is ____.

Correct! Wrong!

The frequency, ν = Rc (1/n12−1/n22)
We know that the Rydberg constant, R = 1.097 ×107m−1
c = 3 × 1088 ms−1
Given, n1=2 and n2=3
Then the frequency, ν = 1.097 × 107×3×108× (1/22−1/32)
= 0.457 × 1015 HZ
= 4.57 × 1014Hz

Ionisation energy of a He+ atom in its ground state is____.

Correct! Wrong!

Ionisation energy is the energy required to remove the electron from its ground state. Which is equal to the energy of the electron in the ground state.

In alpha-particle scattering, the alpha particles that comes closer to the nuclei are____.

Correct! Wrong!

Alpha particles are doubly ionised helium atoms. The charge of the alpha particles is positive. When these alpha particles come closer to the nucleus, they deflected more due to the positive charge of the nucleus.

In an α-particle experiment, the perpendicular distance between the direction of inital velocity of the α-particle and the gold nucleus is called ____.

Correct! Wrong!

In the α-particle experiment, the perpendicular distance between the direction of initial velocity of the α-particle and the gold nucleus is called impact parameter

As the principal quantum number increases, the orbiting velocity of the electron ____.

Correct! Wrong!

The orbiting velocity of the electron, vn= (1/n)×e2/2hϵ0
where n is the principal quantum number.
e, h and ϵ0 are constants.
Then vn∝1/n
Thus the principal quantum number increases the orbiting velocity of the electron decreases.

The shortest wavelength of Brackett series in hydrogen is ____.

Correct! Wrong!

The empirical formula for Brackett series is,
1/λ=R (1/42−1/n2)
The shortest wavelength is obtained for limiting wavelength of the series, that is n = ∞
We know that the R = 1.097 ×107m−1
Then,
1/λ=1.097×107(1/42−1/∞2)
we know that, 1/∞=0
Then, 1/λ=1.097×107×(1/16)
λ = 14.585 ×10−7m
λ = 14,585 A

The concept of nucleus of an atom was proposed by ____.

Correct! Wrong!

After studying the results in alpha particle scattering experiment, Rutherford proposed that the most of the atom is empty. The entire positive charge of the atom and most of the mass of the atom is concentrated in a small volume called, nucleus.

The spectral series of the hydrogen atom that lies in the visible region of the electromagnetic spectrum is ____.

Correct! Wrong!

The wavelength of the Hα line in the Balmer series is 6563A. Which comes under visible light of the electromagnetic spectrum.

The ratio of the wavelengths of the first line of Lyman series and the first line of the Paschen series is ____.

Correct! Wrong!

The empirical formula of Lyman series is, 1/λ=R(1/12−1/n2)
For the first line of the Lyman series, n = 2,
1/λ1=R(1/12−1/n2)
1/λ1=3R/4
The empirical formula of Paschen series is, 1/λ=R(1/32−1/n2)
For the first line of the Paschen series, n = 4,
1/λ2=R(1/32−1/42)
1/λ2=7R/144
Then the ratio of first line of Lyman series to first line of Paschen series is,
λ2 1=3R/4×144/7R
λ12=7/108

The total energy of the electron in an atom is directly proportional to ____.

Correct! Wrong!

The total energy of the electron in an atom is given by, E = −e2/8πϵ0r
Then, E∝1/r

The ratio of energies of hydrogen atom in its first and second excited states is ____.

Correct! Wrong!

The energy of electron of hydrogen atom in the nth orbit is, En=−13.6/n2eV

Class 12th Physics - 12 Atoms MCQs
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Class 12th Physics - 12 Atoms MCQs