Atoms Test

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Atoms Test - 31

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Question 1
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The hyperfine lines in the spectrum is related to

A
Zeeman effect

B
Stark effect

C
Lande's splitting

D
nuclear magnetic spin

Solution

In atomic physics, hyperfine structure is the different effects leading to small shifts and splitting in the energy levels of atoms, molecules and ions. The name is a reference to the fine structure which results from the interaction between the magnetic moments associated with electron spin and the electrons' orbital angular momentum. Hyperfine structure, with energy shifts is typically orders of magnitude smaller than the fine structure, results from the interactions of the nucleus (or nuclei, in molecules) with internally generated electric and magnetic fields.
Question 2
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Directions For Questions

A Geiger counter detects radiation such as Beta particles by using the fact that the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it (34.17). A large potential difference is established between the wire and outer cylinder, with the wire at higher potential, this sets up a strong electric field directed radially outward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus, a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audible "click". Suppose the radius of the central wire is $$145 \mu m$$ and the radius of the hollow cylinder is $$1.80cm$$

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The potential difference between the wire and the cylinder produces an electric field of $$2\times10^{4} Vm^{-1}$$ at a distance of $$1.2 cm$$ from the wire

A
$$360 V$$

B
$$180 V$$

C
$$200 V$$

D
$$240 V$$

Solution

$$E$$ = $$\displaystyle\ \frac{V}{d}$$

$$V$$ $$=Ed$$ $$=2\times10^{4}\times1.2\times10^{-2}$$ $$ = 240 V$$
Question 3
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Which of the following does not fit into a group

A
Photon

B
Graviton

C
Proton

D
Meson

Solution

Graviton, proton and meson are subatomic particles.
But photon is light energy particle.
So, the answer is option (A).
Question 4
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Suppose a particle $$a$$ of charge $$2e$$ and mass $$2m_{e}$$ ($$m_{e}$$ being mass of electron) revolves around the proton in nearest allowed circular orbit of radius r. If the radius of first Bohr orbit for the electron is $$A_{0}$$, then the relation between $$r$$ and $$a_{0}$$ is

A
$$a_{0}$$ = $$4r$$

B
$$a_{0}$$ =$$8r$$

C
$$a_{0}$$ = $$2r$$

D
$$a_{0}$$ = $$\displaystyle\ \frac{r}{4}$$

Solution

$$r=a_o\dfrac{n^2}{Z^2}$$

Here $$n=1 ,Z=2$$

This gives us $$a_o=4r$$
Question 5
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According to Geiger-Nuttal Law the curve between $$log \lambda$$ and $$log R$$ will be

A

B

C

D

Solution

The Geiger–Marsden experiment(s) (also called the Rutherford gold foil experiment) were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where its positive charge and most of its mass are concentrated.They deduced this by measuring how an alpha particle beam is scattered when it strikes a thin metal foil.
At Rutherford's behest, Geiger and Marsden performed a series of experiments where they pointed a beam of alpha particles at a thin foil of metal and measured the scattering pattern by using a fluorescent screen. They spotted alpha particles bouncing off the metal foil in all directions, some right back at the source. This should have been impossible according to Thomson's model; the alpha particles should have all gone straight through. Obviously, those particles had encountered an electrostatic force far greater than Thomson's model suggested they would, which in turn implied that the atom's positive charge was concentrated in a much tinier volume than Thomson imagined.

When Geiger and Marsden shot alpha particles at their metal foils, they noticed only a tiny fraction of the alpha particles were deflected by more than 90°. Most just flew straight through the foil. This suggested that those tiny spheres of intense positive charge were separated by vast gulfs of empty space.Most particles passed through the empty space and experienced negligible deviation, while a handful struck the nuclei of the atoms and bounced right back.

Rutherford thus rejected Thomson's model of the atom, and instead proposed a model where the atom consisted of mostly empty space, with all its positive charge concentrated in its center in a very tiny volume, surrounded by a cloud of electrons.

So the relation obtained is :- logλ=mlogR+D
above equation represent a straight line equation with positive slope m and intercept D with positive x axis.
Question 6
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When $$Z$$ is doubled in an atom, which of the following statement is consistent with Bohr's theory?

A
Energy of a state is doubled

B
Radius of an orbit is doubled

C
Velocity of electrons in the orbit is doubled

D
Radius of an orbit is quadrupled.

Solution

Let the initial charge be $$z$$.
$$\dfrac{kz^2}{r^2}=\dfrac{m_ev^2}{r}$$

when $$z$$ is doubled,

$$\dfrac{k{(2 z)}^2}{r^2}=\dfrac{m_e(v_1)^2}{r}$$

$$v_1=2v$$
Question 7
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The ratio of radii of first orbit of hydrogen atom and the second orbit of singly ionised helium atom will be

A
$$1:2$$

B
$$4:1$$

C
$$1:4$$

D
$$8:1$$

Solution

For hydrogen and hydrogen like atoms,
$$r_n \propto \dfrac{n^2}{Z}$$
where, $$r_n$$ is the radius of $$n^{th}$$ orbit of revolution of electron, Z is the atomic number.
For hydrogen atom, Z = 1. Hence, radius of first orbit is: $$r_h = \dfrac{1^2}{1} = 1$$
Also, or helium atom, Z = 2. Hence, radius of second orbit is: $$r_{he} = \dfrac{2^2}{2} = 2$$
Therefore, $$\dfrac{r_h}{r_{he}} = \dfrac{1}{2}$$
Question 8
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Hydrogen atoms are excited from ground state to the state of principal quantum number $$4$$. Then, the number of spectral lines observed will be

A
$$3$$

B
$$6$$

C
$$5$$

D
$$2$$

Solution

HINT: Use hydrogen spectrum line formulae

STEP 1: Number of spectral lines

Given that quantum number = 4

No. of spectral lines = $$\dfrac{n(n-1)}2$$

where $$n =$$ quantum number

= $$\dfrac{4(4-1)}2\;\;=\;\;\dfrac{4(3)}2\;\;=\;\;6$$

Thus, option B is correct.
Question 9
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What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $$-3.4\space eV$$?

A
$$L = \displaystyle\frac{h}{\pi}$$

B
$$L = \displaystyle\frac{4h}{\pi}$$

C
$$L = \displaystyle\frac{3h}{\pi}$$

D
$$L = \displaystyle\frac{5h}{\pi}$$

Solution

First, we need to identify the quantum number of the energy level.
As $$E = - \displaystyle\frac{13.6}{n^2}$$, we have
$$\quad -3.4 = -\displaystyle\frac{13.6}{n^2}$$ or $$n = 2$$
The angular momentum quantization gives, $$L = mvr = \displaystyle\frac{nh}{2\pi}$$
Substituting $$n=2$$, we get $$L = \displaystyle\frac{2h}{2\pi} = \displaystyle\frac{h}{\pi}$$
Question 10
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There are seven orbitals in a subshell then the value of $$l$$ for it will be

A
$$l=4$$

B
$$l=3$$

C
$$l=2$$

D
$$l=1$$

Solution

The shape of orbitals i.e. number of orbitals depends on the subshell in which they are found.
The maximum possible number of orbitals i.e. the allowed orientations in space are denoted by magnetic quantum number and is given by
$$m_l = 2l +1$$
$$7 = 2l +1$$
$$2l = 6$$
$$l = 3$$