Structure of Atom Test

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Structure of Atom Test - 56

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Weekly Quiz Competition

Question 1
1 / -0

Potassium selenate is isomorphous with the potassium sulphate and contains 50.0% of $$Se$$. Find the atomic weight of $$Se$$.

A
142

B
71

C
47.33

D
284

Solution

Pottassium selenate is isomorphous to $$K_2SO_4$$ and thus its molecular formula is $$K_2SeO_4$$

Now mol.wt of $$K_2SeO_4=(39\times 2+a+4\times 16)$$

$$\Rightarrow 142+a$$

Where $$'a'$$ is at.wt of $$Se$$

$$(142+a)g\;K_2SeO_4$$ $$=Se\ a\ g$$

100g $$K_2SeO_4=\dfrac{a\times 100}{142+a}$$

$$\%$$ of $$Se=50$$

$$a=142$$

$$\approx 142$$

Also equivalent weight of $$K_2SeO_4=\dfrac{Mol.wt}{2}$$

$$\Rightarrow \dfrac{2\times 39+142+64}{2}$$

$$\Rightarrow 142$$

Hence $$(A)$$ is the correct answer.
Question 2
1 / -0

The radius of first excited state of $${ Be }^{ 3+ }$$ ion is:

A
$$0.53\overset { o }{ A } $$

B
$$\cfrac { 0.53 }{ 4 } \overset { o }{ A } $$

C
$$0.53\times 4\overset { o }{ A } $$

D
$$0.53\times 2\overset { o }{ A } $$

Solution
Question 3
1 / -0

The e/m of proton is

A
$$1.78\times {10}^{8}C/g$$

B
$$9.57\times {10}^{4}C/g$$

C
$$19.14\times {10}^{4}C/g$$

D
$$0.478\times {10}^{4}C/g$$

Solution
Question 4
1 / -0

The moment of momentum of electrons in the third orbit in a hydrogen atom

A
$$31.67 \times 10^{-34} gm \ m^2/s$$

B
$$3.167 \times 10^{-34} kg \ m^2/s$$

C
$$31.67 \times 10^{-34} kg \ cm^2/s$$

D
$$31.67 \times 10^{-34} gm \ cm^2/s$$

Solution
Question 5
1 / -0

The radius of the hydrogen atom in the ground state is $$0.53 {\mathring A}$$, the radius of $$L{i^{2 + }}$$ in a similar state is____.

A
$$1.06 {\mathring A}$$

B
$$0.265 {\mathring A}$$

C
$$0.175 {\mathring A}$$

D
$$0.53 {\mathring A}$$

Solution

Radius of hydrogen like particle $$r_n=a_0\dfrac{n^2}{2}$$
$$0.530 \mathring { A }=a_0\dfrac{1^2}{1}$$
$$a_0=0.53\mathring{A}$$
for $$Li^{2+}$$ $$r=a_0\dfrac{1^2}{3}$$
$$=\dfrac{0.530}{3}=0.176 \mathring{A}$$
Hence, option $$C$$ is correct.
Question 6
1 / -0

Ionization energy of $${He}^{+}$$ is $$19.6\times {10}^{-18}J{atom}^{-1}$$. The energy of the first stationary state ($$n=1$$) of $${Li}^{2+}$$ is

A
$$4.41\times {10}^{-16}J{atom}^{-1}$$

B
$$-4.41\times {10}^{-17}J{atom}^{-1}$$

C
$$-2.2\times {10}^{-15}J{atom}^{-1}$$

D
$$-8.82\times {10}^{-17}J{atom}^{-1}$$

Solution

$$ IE$$ of $$ { He }^{ + }$$=$$19.6\times { 10 }^{ -18 } J$$ per atom
$$ IE\quad =\quad { E }_{ 1\quad }(for\quad H)\times { Z }^{ 2 }$$
$${ E }_{ 1 }\times 4= -19.6\times { 10 }^{ -18 }\quad J$$
$${ E }_{ 1 }(for \; {Li}^{+2})={ E }_{ 1 }(for\quad H)\times 9$$
=$$- \cfrac { 19.6\times { 10 }^{ -18 } }{ 4 } \times 9$$
=$$- 441\times { 10 }^{ -18 }$$
=$$- 4.41\times { 10 }^{ -17 } J$$per atom
Question 7
1 / -0

Which among the following elements does not contain same number of s-electrons as the number of d-electrons in $${ Ni }^{ 2+ }$$? (Atomic number of Kr is 36)

A
Ca

B
Kr

C
Cu

D
Fe

Solution
Question 8
1 / -0

Name the compounds used in preparation of nylon-66.

A
E-caprolactum

B
Hexamethylenediamine and adipic acid

C
Dimethyl tetraphthalate

D
Hexamethylene diamine

Solution

Nylon 66 is made of two monomers each containing 6 carbon atoms, hexamethylenediamine and adipic acid, which give nylon 66 its name.
Question 9
1 / -0

The radius of the first Bohr orbit of a hydrogen atom is $$ 0.53\times 10^{-10}m$$.When an electron collides with this atom which is in its normal state, the radius of the electron orbit in the atom change to $$2.12 \times 10^{10}m$$.The value of the principal quantum number $$n$$ of the state to which it is excited is:

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$r_n=a_0n^2$$
$$2.12\times 10^{-19}=90n^2$$
$$=0.53n^2$$
$$n^2=4$$
$$n=2$$
This value of principle quantum number of excited state in $$2$$
Question 10
1 / -0

If transition of an electron in $$H-atom$$ takes place from $$3rd$$ orbit to $$5th$$ then change in angular momentum of electron is given by?

A
$$h$$

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

C
$$2h$$

D
$$3h$$

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Structure of Atom Test - 56

Result

Structure of Atom Test - 56

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SHARING IS CARING

Question 1

142

71

47.33

284

Solution

Question 2

$$0.53\overset { o }{ A } $$

$$\cfrac { 0.53 }{ 4 } \overset { o }{ A } $$

$$0.53\times 4\overset { o }{ A } $$

$$0.53\times 2\overset { o }{ A } $$

Solution

Question 3

$$1.78\times {10}^{8}C/g$$

$$9.57\times {10}^{4}C/g$$

$$19.14\times {10}^{4}C/g$$

$$0.478\times {10}^{4}C/g$$

Solution

Question 4

$$31.67 \times 10^{-34} gm \ m^2/s$$

$$3.167 \times 10^{-34} kg \ m^2/s$$

$$31.67 \times 10^{-34} kg \ cm^2/s$$

$$31.67 \times 10^{-34} gm \ cm^2/s$$

Solution

Question 5

$$1.06 {\mathring A}$$

$$0.265 {\mathring A}$$

$$0.175 {\mathring A}$$

$$0.53 {\mathring A}$$

Solution

Question 6

$$4.41\times {10}^{-16}J{atom}^{-1}$$

$$-4.41\times {10}^{-17}J{atom}^{-1}$$

$$-2.2\times {10}^{-15}J{atom}^{-1}$$

$$-8.82\times {10}^{-17}J{atom}^{-1}$$

Solution

Question 7

Ca

Kr

Cu

Fe

Solution

Question 8

E-caprolactum

Hexamethylenediamine and adipic acid

Dimethyl tetraphthalate

Hexamethylene diamine

Solution

Question 9

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 10

$$h$$

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

$$2h$$

$$3h$$

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