Structure of Atom Test

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Structure of Atom Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

An electron with energy $$12.09$$ eV strikes hydrogen atom in ground state and gives its all energy to the hydrogen atom. Therefore hydrogen atom is excited to __________ state.

A
Fourth

B
Third

C
Second

D
First

Solution

We know that energy of Hydrogen atom in it's orbit is given by formula $$\dfrac{-13.6}{n^2} eV$$.

Initial Energy of Electron = $$-13.6 eV$$ (Given that atom in i's ground state i.e n=1)

Energy after striking = $$(-13.6 + 12.09) eV = -1.51 eV$$

$$\Rightarrow$$ $$\dfrac{-13.6}{n^2} eV = 1.51eV$$ $$\Rightarrow$$ $$n^2= 9$$ $$\Rightarrow$$ $$ n=3.$$

So, hydrogen atom is excited to third state.

Therefore, B is correct option.
Question 2
1 / -0

Which of the following properties of atom could be explained correctly by Thomson's model of atom?

A
Overall neutrality of atom

B
Spectra of hydrogen atom

C
Position of electrons, protons and neutrons in atom

D
Stability of atom

Solution

Hint: All atoms include tiny negatively charged subatomic particles or electrons, according to J.J. Thomson's work with cathode ray tubes.

Explanation:

According to the Thomson atom model:

1. An atom is spherical and has a radius of $$10^{-10}$$ meters.

2. The positive charge in the atom is evenly distributed.

3. The electrons are embedded in such a way that the atom's electrostatic configuration is the most stable.

As a result, the model was able to explain the atom's overall neutrality.

4. The atom's mass is considered to be evenly distributed across the atom.

It was unable to account for atomic spectra, atomic stability, or the real distribution of electrons, protons, and neutrons within the atom.

Final Answer:

The correct answer is option $$(A)$$.
Question 3
1 / -0

The spectrum of a molecule is so detailed that a molecule can be identified by its spectrum.

A
Only if it is small

B
Only by test-tube analysis

C
With great difficulty

D
Easily, like a fingerprint

Solution

A substance can be identified because it's chemical enviroment often shows up like a finger print in a spectrum .This fine region of finger prints of the chemical enviroment of a decimal compound is created by infrared radiation
Question 4
1 / -0

One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in:

A
visible region

B
infrared region

C
ultraviolet region

D
microwave region

Solution

$$\begin{array}{l}h\nu = 1.6 \times {10^{ - 19}}\\\nu = \dfrac{{1.6 \times {{10}^{ - 19}}}}{{6.626 \times {{10}^{ - 34}}}} = 2.4 \times {10^{14}}Hz\end{array}$$
Question 5
1 / -0

Consider a $$p_{y}$$-orbital of an atom and identify correct statement.

A
$$s-$$orbital of another atom produces $$\pi$$-bond when y is the bond formation axis.

B
$$p_{y}$$-orbital of another atom produces $$\Sigma$$-bond when x is the bond formation axis.

C
$$p_{z}$$-orbital of another atom produces $$\Sigma$$-bond when x is the bond formation axis.

D
$$d_{xy}$$-orbital of another atom produces $$\Sigma$$-bond when x is the bond formation axis.

Solution
Question 6
1 / -0

According to Bohr's theory, the angular momentum for an electron of $${5}^{th}$$ orbit is:

A
$$2.5{h}/{\pi}$$

B
$$5{h}/{\pi}$$

C
$$25{h}/{\pi}$$

D
$$6{h}/{2\pi}$$

Solution

$$\bf{Explanation-}$$
According to Bohr's theory the angular momentum of an electron is given as;
$$mvr = \dfrac{nh}{2\pi}$$

For $$5th$$ orbit, $$n = 5$$

Putting the value of $$n$$ in the above equation we get;
Angular momentum $$= \dfrac{5h}{2\pi} = \dfrac{2.5h}{\pi}$$

$$\bf{Conclusion-}$$ Hence, option A is correct.
Question 7
1 / -0

In a Bohr's model of an atom when an electron jumps from n=1 to n=3, how much energy will be emitted or absorbed in erg?

A
$$2.15\times { 10 }^{ -11 }$$

B
$$0.191\times { 10 }^{ -10 }$$

C
$$2.389\times { 10 }^{ -12 }$$

D
$$0.239\times { 10 }^{ -10 }$$

Solution
Question 8
1 / -0

The number of orbital having values of n = 4 and l = 3 :

A
3

B
7

C
5

D
9

Solution

The number of orbital having values of $$n=4$$ and$$ l=3$$
$$\Rightarrow 4f$$
f subshell
7 orbitals
$$\begin{bmatrix} f-value & 0 & 1 & 2 & 3 \\ orbital & s & p & d & f \end{bmatrix}$$
Question 9
1 / -0

Which of the following contains maximum number of atoms?

A
$$0.16\ g\ CH_3$$

B
$$0.28\ g \ N_2$$

C
$$0.36\ g\ C_6H_{12}O_6$$

D
$$1.1\ g \ CO_2$$

Solution

No. of atoms in $$CH_3$$ = $$N_a \times mole$$
= $$N_a \times \dfrac{0.16}{15}$$

No. of atoms in $$N_2$$ =$$N_a \times mole$$
=$$N_a \times \dfrac{0.28}{28}$$

No. of atoms in $$C_6H_{12}O_6 = N_a \times \dfrac{0.36}{180}$$

No. of atoms in $$CO_2 = N_a\times \dfrac{1.1}{44}$$
Hence $$1.1$$ g of $$CO_2$$ has the maximum no. of atoms.
Thus D is the correct option.
Question 10
1 / -0

Atomic Radius is of the order $$10^{-8}$$cm & nucleus order of $$10^{-13}$$cm. Calculate what fraction of atom is occupied by nucleus?

A
$$10^{-14} \times V_{atom}$$

B
$$10^{-13} \times V_{atom}$$

C
$$10^{-15} \times V_{atom}$$

D
$$10^{-11} \times V_{atom}$$

Solution

We know the volume of a sphere varies to 3rd power of the radius and also nucleus and atoms are spherical in shape

$$\dfrac{V_{atom}}{V_{nucleus}}=\dfrac{(10^{-8})^3}{(10^{-13})^3}$$

$$\dfrac{V_{atom}}{V_{nucleus}}=\dfrac{1}{(10^{-5})^3}$$

$$V_{nucleus} = 10^{-15}\times V_{atom}$$

Hence, the correct option is $$C$$

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Structure of Atom Test - 50

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Structure of Atom Test - 50

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SHARING IS CARING

Question 1

Fourth

Third

Second

First

Solution

Question 2

Overall neutrality of atom

Spectra of hydrogen atom

Position of electrons, protons and neutrons in atom

Stability of atom

Solution

Question 3

Only if it is small

Only by test-tube analysis

With great difficulty

Easily, like a fingerprint

Solution

Question 4

visible region

infrared region

ultraviolet region

microwave region

Solution

Question 5

$$s-$$orbital of another atom produces $$\pi$$-bond when y is the bond formation axis.

$$p_{y}$$-orbital of another atom produces $$\Sigma$$-bond when x is the bond formation axis.

$$p_{z}$$-orbital of another atom produces $$\Sigma$$-bond when x is the bond formation axis.

$$d_{xy}$$-orbital of another atom produces $$\Sigma$$-bond when x is the bond formation axis.

Solution

Question 6

$$2.5{h}/{\pi}$$

$$5{h}/{\pi}$$

$$25{h}/{\pi}$$

$$6{h}/{2\pi}$$

Solution

Question 7

$$2.15\times { 10 }^{ -11 }$$

$$0.191\times { 10 }^{ -10 }$$

$$2.389\times { 10 }^{ -12 }$$

$$0.239\times { 10 }^{ -10 }$$

Solution

Question 8

3

7

5

9

Solution

Question 9

$$0.16\ g\ CH_3$$

$$0.28\ g \ N_2$$

$$0.36\ g\ C_6H_{12}O_6$$

$$1.1\ g \ CO_2$$

Solution

Question 10

$$10^{-14} \times V_{atom}$$

$$10^{-13} \times V_{atom}$$

$$10^{-15} \times V_{atom}$$

$$10^{-11} \times V_{atom}$$

Solution

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