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

The empirical formula of Lyman series is, 1/λ=R (1/1^{2}−1/n^{2})

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(1/2^{2}−1/n^{2})

The first line of the Balmer series is obtained for n = 3.

Then 1/λ_{1}=R (1/2^{2}−1/3^{2})

1/λ_{1}=5R/36

λ_{1}=36/5R

λ_{1}/λ=(36/5R)×(3R/4)

λ_{1}/λ =27/5

λ_{1}=(27/5)λ

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

We know that the Bohr's radius = (0.529×10^{−10}) ×n^{2}m

For the first orbit, n = 1.

Then the Bohr's radius is, r_{1}= (0.529×10^{−10}) ×1^{2}m

For the second orbit, n = 2.

Then the Bohr's radius is, r_{2}= (0.529×10^{−10}) ×2^{2}m

Then the ratio of radii of the first orbit to the second orbit is r_{1}/r_{2}=1/4

r_{1}: r_{2}=1:4

#### 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 ____.

We know that,

The kinetic energy of electron in hydrogen atom, K =e^{2}/8πϵ_{0}r

The total energy of the electron in hydrogen atom, E =−e^{2}/8πϵ_{0}r

On comparing the above two equations, E=−K

#### The shortest wavelength of Brackett series in hydrogen is ____.

The empirical formula for Brackett series is,

1/λ=R (1/4^{2}−1/n^{2})

The shortest wavelength is obtained for limiting wavelength of the series, that is n = ∞

We know that the R = 1.097 ×10^{7}m^{−1}

Then,

1/λ=1.097×10^{7}(1/4^{2}−1/∞^{2})

we know that, 1/∞=0

Then, 1/λ=1.097×10^{7}×(1/16)

λ = 14.585 ×10^{−7}m

λ = 14,585 A

#### An electron jumps from3^{rd} orbit to the 2^{nd}orbit of hydrogen atom. The frequency of the emitted radiation is ____.

The frequency, ν = Rc (1/n_{1}^{2}−1/n_{2}^{2})

We know that the Rydberg constant, R = 1.097 ×10^{7}m^{−1}

c = 3 × 10^{8}8 ms^{−1}

Given, n_{1}=2 and n_{2}=3

Then the frequency, ν = 1.097 × 10^{7}×3×10^{8}× (1/2^{2}−1/3^{2})

= 0.457 × 10^{15} HZ

= 4.57 × 10^{14}Hz

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

According to Bohr's postulate, the angular momentum of a orbiting electron in the nth orbit is, L_{n}=nh/2π

Angular momentum of the electron in the 3^{rd} orbit is, L_{3}=3h/2π

Angular momentum of the electron in the 4^{th} orbit is, L_{4}=4h/2π

Then the difference in the angular momentum is L_{4}−L_{3}= (4h/2π)−(3h/2π)

=h/2π

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

Wave number, 1/λ=R(1/n_{1}^{2}−1/n_{2}^{2})

Given, n_{1}=2 and n_{2}=4

Then, 1/λ=R(1/2^{2}−1/4^{2})

1/λ=R×3/16

Thus the wave number is 3R/16.

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

The orbiting velocity of the electron, v_{n}= (1/n)×e^{2}/2hϵ_{0}

where n is the principal quantum number.

e, h and ϵ_{0} are constants.

Then v_{n}∝1/n

Thus the principal quantum number increases the orbiting velocity of the electron decreases.

#### Ionisation energy of a He^{+} atom in its ground state is____.

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.

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

The empirical formula of Lyman series is, 1/λ=R(1/1^{2}−1/n^{2})

For the first line of the Lyman series, n = 2,

1/λ_{1}=R(1/1^{2}−1/n^{2})

1/λ_{1}=3R/4

The empirical formula of Paschen series is, 1/λ=R(1/3^{2}−1/n^{2})

For the first line of the Paschen series, n = 4,

1/λ_{2}=R(1/3^{2}−1/4^{2})

1/λ_{2}=7R/144

Then the ratio of first line of Lyman series to first line of Paschen series is,

λ_{2 }/λ_{1}=3R/4×144/7R

λ_{1}/λ_{2}=7/108

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

The empirical formula for Balmer series is, 1/λ=R (1/2^{2}−1/n^{2})

The longest wavelength of the series is obtained for n = 3,

1/λ1=R (1/2^{2}−1/3^{2})

1/λ_{1}=5R/36

The shortest wavelength of the series is obtained for n = ∞,

1/λ_{2}=R (1/2^{2}−1/∞^{2})

1/λ_{2}=R/4

Then the ratio of longest wavelength to the shortest wavelength of the series is, λ_{1}/λ_{2}= (36/5R)×(R/4)

λ_{1}/λ_{2}=9/5

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

The energy of electron in the n^{th} orbit of hydrogen like atom of atomic number Z is, E_{n}= (−13.6/n^{2})Z^{2 }eV

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

The wavelength of the H_{α } line in the Balmer series is 6563A. Which comes under visible light of the electromagnetic spectrum.

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

We know that,

Size of an atom is 10^{−10}m

Size of a nucleus is 10^{−14}m

Then the ratio of size of an atom to the size of a nucleus is, 10^{−10}/10^{−14}

=10^{4}:1

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

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.

#### If Rutherford's scattering experiment is conducted with different projectiles of charges q_{1} and q_{2} moving with the same velocity and the masses of the projectiles are M_{1} and M_{2} for the same target. The ratio of the distance of closest approach of the nucleus is ____.

Given,

The charges of the projectiles are q_{1} and q_{2}

The masses of the projectile are M_{1} and M_{2}

At the distance of closest approach,

Kinetic Energy of the projectile = Potential Energy of the system

For the first projectile,

1/2M_{1}v^{2}=1/4πϵ_{0}×[(q_{1}e)(ze)]/d_{1} .... (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,

1/2M_{2}v^{2}=1/4πϵ_{0}× [(q_{2}e)(Ze)]/d_{2 }...(2)

Divide the equations (1) and (2).

Then, M_{1}/M_{2}=q_{1}/q_{2}×d_{2}/d_{1}

M_{1 }/M_{2}×q_{2 }/q_{1}=d_{2}/d_{1}

Thus d_{1}/d_{2}=M_{2}q_{1}/M_{1}q_{2}

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

The energy of electron of hydrogen atom in the n^{th} orbit is, En=−13.6/n^{2}eV

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

The Rutherford's model is unstable according to classical theory

#### The radius of the first orbit of hydrogen atom is called Bohr's radius, a_{0}. The radius of the first orbit of Li^{+} is___.

We know that the Bohr's radius, a_{0}=0.529A

For hydrogen like atoms, the radius of nth orbit is, r=(0.529/Z)n^{2 }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)×1^{2}

=0.529/3 A

=a_{0}/3

#### The longest wavelength of Lyman series for hydrogen is ____.

The empirical formula for Lyman series is

1/λ=R(1/1^{2}−1/n^{2})

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=1.097×10^{7}(1/1^{2}−1/2^{2})

= 1.097 × 10^{7}(1−1/4)

= 0.82275 × 10^{7}m^{−1}

Then λ = 1.215 × 10^{−7}m

= 1215 A

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

Given the kinetic energy of the alpha particle, K.E = 9 MeV

K.E = 9 × 1.6 × 10^{−13}J

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 = 1/4πϵ_{0}×[(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} = (1/4πϵ_{0})×[(2e)(Ze)]/d

we know that, 1/4πϵ_{0}=9×10^{9}Nm^{2}C^{−2}

Then, d = 9×10^{9}×(2×79×1.6×10^{−19}×1.6×10^{−19}/9×1.6×10^{−13})

d = 252.8 × 10^{−16}m

d = 25.28 × 10^{−15}m

d= 25.28 fm

#### According to Bohr's atomic model, the angular momentum of an electron in the n^{th} orbit is ____.

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

The empirical formula for Balmer series is, 1/λ=R(1/2^{2}−1/n^{2})

The second line of the Balmer series is obtained for n = 4

we know that, R=1.097×107m^{−1}

Then 1/λ=1.097×10^{7}(1/2^{2}−1/4^{2})

1/λ=3/16×1.097×10^{7}

λ=4.8617×10^{−7}m

= 4861.7 A

This wavelength represents blue colour of the visible light.

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

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

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

The energy of electron in the nth orbit of hydrogen like atom of atomic number Z is, En=−(13.6/n^{2 })Z^{2}eV

The energy of electron in the first orbit of hydrogen atom

Is, E=− (13.6/1^{2 })1^{2}eV

E=−13.6eV

The energy of electron in the 2nd orbit of He+ atom is, E_{2}=− (13.6/2^{2 }) ×2^{2}eV

E_{2}=−13.6eV

Thus E_{2}=E

#### Plum pudding model of the atom was proposed by ____.

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

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 n^{th} orbit is, En=−13.6/n^{2}eV

The energy of the electron in the ground state is, E_{1}=−13.6/1^{2}

E_{1}=−13.6eV

The energy of the electron in the first excited state is, E_{2}=−13.6/2^{2}

E_{2}=−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.

E_{2} −E_{1} = hν

E_{2} −E_{1} = −3.4 - (−13.6)

E_{2} − E_{1} = 10.2 eV

hν=10.2eV

=10.2×1.6×10^{−19}J

Then the frequency of the radiation, ν=10.2×1.6×10^{−19}/h

=10.2×1.6×10^{−19 }/6.6×10^{−34}Js

=2.4727×10^{15}Hz

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

The total energy of the electron in an atom is given by, E = −e^{2}/8πϵ_{0}r

Then, E∝1/r

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

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.

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