Atoms Test

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Atoms Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

$$_{ 95 }^{ 241 }{ Am }\ \rightarrow \ _{ 93 }^{ 237 }{ Np }+Y$$
In the above nuclear fusion reaction, Identify the particle represented by $$Y$$?

A
A proton

B
An electron

C
An alpha particle

D
A gamma ray

E
A beta particle

Solution

The given nuclear reaction : $$^{241}_{95} Am \rightarrow ^{237}_{93} Np $$ $$+$$ $$^A_Z Y$$
In a nuclear reaction, the mass number as well as the atomic number is conserved.
$$\therefore$$ $$241 = 237 + A$$ $$\implies A =4$$
Also, $$95 = 93 + Z$$ $$\implies Z =2$$
Thus, the particle $$Y$$ is an alpha particle i.e $$^4_2He$$.
Question 2
1 / -0

According to Bohr's model of the atom, atoms emit or absorb radiation only at certain wavelengths.Identify why is it so

A
Because the protons and electrons are distributed evenly throughout the atom

B
Because electrons can orbit the nucleus at any radius

C
Because electrons orbit the nucleus only at certain discrete radii

D
Because protons orbit the nucleus only at certain discrete radii

E
Because photons can only have discrete wavelength

Solution

Bohr's frequency condition is given by $$hc/\lambda=E_{2}-E_{1}$$
Above relation shows that wavelength $$\lambda$$ of radiation depends upon energy difference of two orbits, $$\lambda$$ is certain because energy difference is constant for two orbits it means $$E_{1}$$ and $$E_{2}$$ are also fixed i.e. electron is revolving in some fixed orbits of discrete radii. These orbits are called stationary orbits in which electron does not emit energy when revolving in these orbits.
Question 3
1 / -0

Although Rutherford's planetary model of the atom was a major step forward in understanding the structure of the atom, which of the following was a major problem for this model-one that Bohr's model attempted to fix?

A
Electrons cannot orbit in elliptical paths

B
Cometary electrons would interfere with the normal orbits of electrons

C
Gravity is too strong to allow stable electron orbits

D
Orbiting electrons would emit electromagnetic energy and spiral into the nucleus

E
Orbiting electrons are all the same size, unlike the planets

Solution

A
The motion of the electrons in Rutherford model was unstable because, according to classical mechanics and electromagnetic theory, any charged particle moving on a curved path emits electromagnetic radiation; thus the electrons would lose energy and spiral into nucleus. To remedy the stability problem, Bohr modified the Rutherford model by requiring that the electrons move in a orbit of fixed size and energy. The energy of an electrons depends on the size if the orbit and is lower for smaller orbits. Radiation can occur only when the electrons jumps from one orbit to another. The atom will be completely stable in the state with the smallest orbit. Since there is not orbit of lower energy into which the electron can jump.
Question 4
1 / -0

Total energy of electron in an excited state of hydrogen atom is $$-3.4 eV$$. The kinetic and potential energy of electron in this state

A
$$K=-3.4 eV$$ $$U=-6.8 eV$$

B
$$K=3.4 eV$$ $$U=-6.8 eV$$

C
$$K=-6.8 eV$$ $$U=+3.4 eV$$

D
$$K=+10.2 eV$$ $$U=-13.6 eV$$

Solution

Total energy of electron, $$T = -3.4eV$$ (Given)
Kinetic energy of electron, $$K = |T| =3.4eV$$
Potential energy of electron, $$U = 2T = -6.8eV$$
Question 5
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An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision

A
must be elastic

B
may be partially elastic

C
must be completely inelastic

D
may be completely inelastic.

Solution

In hydrogen atom, $$ E_n = \dfrac{-13.6}{n^2} eV $$
$$ n = 1; E_1 = -13.6 eV $$
$$ n = 2; E_2 = \dfrac{-13.6}{4} eV $$
First exited state = $$ 3.4 eV $$
$$ E_2 – E_1 = 10.2 eV $$

Since energy difference is 10.2eV which is greater than K.E. 5 eV, the electron exites the hydrogen atom, hence the collision must be elastic as there is no loss in energy of the electron.
Question 6
1 / -0

When electron jumps from $$n=4$$ level to $$n=1$$ level, the angular momentum of electron changes by

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

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

C
$$\dfrac{3h}{2\pi}$$

D
$$\dfrac{4h}{2\pi}$$

Solution

Angular momentum of electron in nth state, $$J_n = \dfrac{nh}{2\pi}$$
Change in angular momentum, $$\Delta J = (n_1-n_2)\dfrac{h}{2\pi}$$
$$\therefore$$ $$\Delta J = \dfrac{(4-1)h}{2\pi} = \dfrac{3h}{2\pi}$$
Question 7
1 / -0

Which of the following spectral series of hydrogen atom is lying in visible range of electromagnetic wave?

A
Paschen series

B
Pfund series

C
Lyman series

D
Balmer series

Solution

Lyman series of hydrogen atom lies in ultraviolet region, Balmer series lies in visible region while Pfund and Paschen series lie in infrared region.
Question 8
1 / -0

The packets of light energy are called

A
Electrons

B
Neutrons

C
Photons

D
Positrons

Solution

The packets of light energy are called as photons.
Energy of a photon is given by $$E = h\nu$$
where $$\nu$$ is the frequency of light incident.
Question 9
1 / -0

As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom

A
decreases

B
increases

C
remains the same

D
does not increase.

Solution

Total energy of the electron in nth orbit $$E_n = \dfrac{-13.6}{n^2} \ eV$$
Potential energy $$P.E = 2E_n = \dfrac{-27.2}{n^2} \ eV$$
Thus potential energy of the atom increases with increase in $$n$$.
Question 10
1 / -0

In which of the following transitions will the wavelength be minimum ?

A
$$n = 5$$ to $$n = 4$$

B
$$n = 4$$ to $$n = 3$$

C
$$n = 3$$ to $$n = 2$$

D
$$n = 2$$ to $$n = 1.$$

Solution

Wavelength of emitted photon due to transition :
$$\dfrac{1}{\lambda} = R\bigg(\dfrac{1}{n_2^2}-\dfrac{1}{n_1^2}\bigg)$$
(A) : $$n_1 = 5$$     $$n_2 =4$$
$$\therefore$$ $$\dfrac{1}{\lambda_A} = R\bigg(\dfrac{1}{4^2}-\dfrac{1}{5^2}\bigg)$$
$$\implies \ \lambda_A = \dfrac{44.44}{R}$$
(B) : $$n_1 = 4$$     $$n_2 =3$$
$$\therefore$$ $$\dfrac{1}{\lambda_B} = R\bigg(\dfrac{1}{3^2}-\dfrac{1}{4^2}\bigg)$$
$$\implies \ \lambda_B = \dfrac{20.57}{R}$$
(C) : $$n_1 = 3$$     $$n_2 =2$$
$$\therefore$$ $$\dfrac{1}{\lambda_C} = R\bigg(\dfrac{1}{2^2}-\dfrac{1}{3^2}\bigg)$$
$$\implies \ \lambda_C = \dfrac{7.2}{R}$$
(D) : $$n_1 = 2$$     $$n_2 =1$$
$$\therefore$$ $$\dfrac{1}{\lambda_D }= R\bigg(\dfrac{1}{1^2}-\dfrac{1}{2^2}\bigg)$$
$$\implies \ \lambda_D = \dfrac{1.33}{R}$$
Thus wavelength is minimum in $$n=2$$ to $$n=1$$.

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Atoms Test - 40

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Atoms Test - 40

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Question 1

A proton

An electron

An alpha particle

A gamma ray

A beta particle

Solution

Question 2

Because the protons and electrons are distributed evenly throughout the atom

Because electrons can orbit the nucleus at any radius

Because electrons orbit the nucleus only at certain discrete radii

Because protons orbit the nucleus only at certain discrete radii

Because photons can only have discrete wavelength

Solution

Question 3

Electrons cannot orbit in elliptical paths

Cometary electrons would interfere with the normal orbits of electrons

Gravity is too strong to allow stable electron orbits

Orbiting electrons would emit electromagnetic energy and spiral into the nucleus

Orbiting electrons are all the same size, unlike the planets

Solution

Question 4

$$K=-3.4 eV$$ $$U=-6.8 eV$$

$$K=3.4 eV$$ $$U=-6.8 eV$$

$$K=-6.8 eV$$ $$U=+3.4 eV$$

$$K=+10.2 eV$$ $$U=-13.6 eV$$

Solution

Question 5

must be elastic

may be partially elastic

must be completely inelastic

may be completely inelastic.

Solution

Question 6

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

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

$$\dfrac{3h}{2\pi}$$

$$\dfrac{4h}{2\pi}$$

Solution

Question 7

Paschen series

Pfund series

Lyman series

Balmer series

Solution

Question 8

Electrons

Neutrons

Photons

Positrons

Solution

Question 9

decreases

increases

remains the same

does not increase.

Solution

Question 10

$$n = 5$$ to $$n = 4$$

$$n = 4$$ to $$n = 3$$

$$n = 3$$ to $$n = 2$$

$$n = 2$$ to $$n = 1.$$

Solution

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