Structure of Atom Test

Result

Structure of Atom Test - 55

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

According to Bohr's atomic theory, which of the following is correct?

A
Potential energy of electrom $$\alpha $$$$\dfrac{Z^2}{n^2}$$

B
The product of velocity of electron and principle quantum number (n) $$\alpha Z^2$$

C
Frequency of revolution of electron in an orbit $$\alpha \dfrac{Z^2}{n^3}$$

D
Coulombic force of attraction on the electron $$\alpha \dfrac{Z^2}{n^2}$$

Solution

Potential energy of an electron is given by,
$$P.E=-\dfrac{2\pi^2 mZ^2 e^4}{n^2h^2}$$
Therefore, $$P.E$$ $$ \alpha$$ $$\dfrac{Z^2}{n^2}$$
velocity of an electron is given by,
$$v=\dfrac{2\pi Ze^2}{nh}$$
Therefore,
$$(v\times n)$$ $$\alpha$$ $$Z$$
Frequency of an electron is given by,
$$f=\dfrac{4\pi^2 Z^2 e^4}{n^3 h^3}$$
$$f$$ $$\alpha$$ $$\dfrac{Z^2}{n^3}$$
Coloumbic force of attraction is given by,
Coloumbic force of attraction of an electron $$\alpha$$ $$\dfrac{Z^3}{n^4}$$
Question 2
1 / -0

What atomic number of an element $$X$$ would have to become so that the $$4th$$ orbit around $$X$$ would fit inside the $$1st$$ Bohr orbit of $$H-$$atom?

A
3

B
4

C
16

D
25

Solution

By definition of Bohr radius $$a_o$$,
$$r=n^2\left(\cfrac {a_o}{Z}\right), r=$$ radical distance from the nucleus of a $$H$$-like atom
So, if we want $$r=a_o$$,
$$Z=n^2\left(\cfrac {a_o}{r}\right)=n^2\left(\cfrac {a_o}{a_o}\right)=n^2$$
For the $$4^{th}$$ Bohr orbit, $$n=4$$,
$$\Rightarrow Z= 4^2= 16$$
Question 3
1 / -0

What is the seperation energy (in eV) for $$Be^{3+}$$ in the first excited state in eV?

A
$$13.6eV$$

B
$$27.2eV$$

C
$$40.8eV$$

D
$$54.4eV$$

Solution

Separation Energy= $$13.6 \times \cfrac {Z^2}{n^2}$$
For $$Be$$, $$Z=4$$
For $$1^{st}$$ excited state $$n=2$$
$$\therefore 13.6 \times \cfrac {4^2}{2^2}= 54.4eV$$
Question 4
1 / -0

For Hydrogen atom the ratio of time required to complete $$1$$ revolution of electron in $$I^{st},II^{nd}$$ and $$III^{rd}$$ orbit respectively is:

A
$$1:8:27$$

B
$$1:4:9$$

C
$$1:2:3$$

D
$$1:16:81$$

Solution

Time required to complete $$1$$ revolution $$=\cfrac{2\pi r}{v}$$
$$\because r\propto n^2$$ and $$v\propto \cfrac{1}{n}$$
$$\therefore $$ $$Time$$ $$required\propto n^3$$
$$\therefore$$Ratio of time required in $$I^{st},II^{nd}$$ and $$III^{rd}$$ orbit is, $$(1)^3:(2)^3:(3)^3$$ i.e. $$1:8:27$$
Question 5
1 / -0

The sulphate of a metal $$M$$ contains $$9.87\%$$ of $$M$$. This sulphate is isomorphous with $$ZnSO_{4}.7H_{2}O$$. The atomic weight of $$M$$ is:

A
$$40.3$$

B
$$36.3$$

C
$$24.3$$

D
$$9.87$$

Solution

1) Consider the molecular weight of the metal be M

Now as it is isomorphous with $$ZnSO_4.7H_2O$$ it would be of the formula $$MSO_4.7H_2O$$

Percentage of M in the compound would be

$$M \times \cfrac{100}{M}+$$ weight of $$SO_4+7\times $$ weight of $$H_2O$$

$$=\cfrac{M\times 100}{M+96+126}$$

$$=9.87$$
Question 6
1 / -0

Which of the following frequencies is associated with the visible region of a spectrum?

A
$$3\times10^9$$ cycle /seconds

B
$$6\times10^{14}$$ cm cycle /seconds

C
$$5.0\times10^{10-3}$$ cycle /seconds

D
None of these

Solution

Frequency range of visible region $$430-770$$
$$TH_{2}$$
$$430\times 10^{12}-770\times 10^{12}Hz$$
$$6\times 10^{14}$$ cycle/second associated with visible region
Question 7
1 / -0

The percentage of an element $$M$$ is $$53$$ percentage in its oxide of molecular formula $${ M }_{ 2 }{ O }_{ 3 }$$. Its atomic mass is about:

A
45

B
9

C
18

D
38

E
27

Solution

Assume atomic mass of $$M$$ is $$x$$
Molecular weight of compound $$=2x + 16\times 3$$
$$=2x + 48$$
% of weight of M $$=\dfrac{2x}{2x+48}\times 100=53(given)$$
Solve for $$x$$,
$$x=27.06=27$$
Question 8
1 / -0

Atomic number of the central atom in $${ MCl }_{ 2 }$$ is $$50$$. The shape of gaseous $${ MCl }_{ 2 }$$ is given as:

A

B

C

D

Solution

$$MC{{L}_{2}}$$ have $$ 50$$ atomic number so $$ M$$ will be $$ 16$$ atomic number.
$$ C{{l}_{2}}=17+17 $$
atomic number is $$34+16=50$$.
And molecule will be $$SC{{l}_{2}}$$ and according to this shape of the molecule is bent.

And $$S$$ have two lone pair.
Question 9
1 / -0

The ratio of the radius of $$1st$$ orbit of $$H$$ atom to that of its $$4$$th excited state is:

A
$$\dfrac{1}{5}$$

B
$$\dfrac{1}{25}$$

C
$$25$$

D
$$\dfrac{1}{16}$$

Solution

Bohr radius r $$\propto \dfrac{n^{2}}{2}$$

$$\dfrac{r_{1st} \, orbit }{r _{4th} \, excited \, state} = \dfrac{1^{2}}{4^{2}}$$

= $$\dfrac{1}{16}$$

Option D is the correct answer.
Question 10
1 / -0

The atomic masses of two elements A and B are 20 and 40 respectively. If x gm of A contains Y atoms, how many atoms are present in 2x gm of B?

A
$$2y$$

B
$$y/2$$

C
$$y$$

D
$$4y$$

Solution

According to Avogadro's law, $$1$$ mole of every substance weighs equal to the molecular mass and contains Avogadro's number, $$6.023 \times 10^{23} $$ of particles.

Moles $$= \dfrac{Given \ mass}{Molar \ mass}$$

Moles of $$A = \dfrac{x}{20} = 0.05 \times x moles$$

Thus, $$0.05 \times x$$ moles of A contain $$= y$$ atoms

Moles of $$B = \dfrac{2x}{40} = 0.05 \ x$$ moles

Thus $$0.05 \times x$$ moles of B contains $$= y$$ atoms

As equal mole contains equal atoms.

Thus, the correct answer is $$'y'$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 55

Result

Structure of Atom Test - 55

Score

-

Rank

-

SHARING IS CARING

Question 1

Potential energy of electrom $$\alpha $$$$\dfrac{Z^2}{n^2}$$

The product of velocity of electron and principle quantum number (n) $$\alpha Z^2$$

Frequency of revolution of electron in an orbit $$\alpha \dfrac{Z^2}{n^3}$$

Coulombic force of attraction on the electron $$\alpha \dfrac{Z^2}{n^2}$$

Solution

Question 2

3

4

16

25

Solution

Question 3

$$13.6eV$$

$$27.2eV$$

$$40.8eV$$

$$54.4eV$$

Solution

Question 4

$$1:8:27$$

$$1:4:9$$

$$1:2:3$$

$$1:16:81$$

Solution

Question 5

$$40.3$$

$$36.3$$

$$24.3$$

$$9.87$$

Solution

Question 6

$$3\times10^9$$ cycle /seconds

$$6\times10^{14}$$ cm cycle /seconds

$$5.0\times10^{10-3}$$ cycle /seconds

None of these

Solution

Question 7

45

9

18

38

27

Solution

Question 8

Solution

Question 9

$$\dfrac{1}{5}$$

$$\dfrac{1}{25}$$

$$25$$

$$\dfrac{1}{16}$$

Solution

Question 10

$$2y$$

$$y/2$$

$$y$$

$$4y$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.