Atoms Test

Result

Atoms Test - 57

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

The distance between $$3^{rd}$$ & $$2^{nd}$$ Bohr orbit $$He^{+}$$ is

A
$$2.645\ A^o$$

B
$$1.322\ A^o$$

C
$$0.2645\ A^{o}$$

D
$$None\ of\ these$$

Solution

Dis\tance between the $${ 2^{ nd } }$$ and the $${ 3^{ rd } }$$ orbits=distance of $${ 3^{ rd } }$$ orbit-distance of second orbit from the nucleus
$${ d_{ 3 } } = 0.529 \times \dfrac { n^{ 2 }}{ z } \times { 10^{ -10 } } m=0.529 \times 9 \times { 10^{ -10 } } m $$
$${ d_{ 2 } }=0.529\times 4\times { 10^{ -10 } }\, m \\ { d_{ 3 } }-{ d_{ 2 } }=0.529\times 5\times { 10^{ -10 } }\, m =2.645\times { 10^{ -10 } }\, m=2.645A^{ 0 } $$

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

The ratio of the radii of the first three Bohr orbit in H atom is

A
$$1:\dfrac { 1 }{ 2 } :\dfrac { 1 }{ 3 } $$

B
$$1:2:3$$

C
$$1:4:9$$

D
$$1:8:27$$

Solution

Atomic number $$Z$$ is equal to $$1$$.
Hence the radius of $$n th$$ orbit, $$r_n=0.529 n^2 A$$
For first three orbits, $$n$$ values are $$1,2$$ and $$3$$
ratio of radii of first three orbit $${r_1}:{r_2}:{r_3} = n_1^2:n_2^2:n_3^2$$
$$\begin{array}{l} \Rightarrow { 1^{ 2 } }:{ 2^{ 2 } }:{ 3^{ 2 } } \\ \Rightarrow 1:4:9 \end{array}$$
Question 3
1 / -0

The angular momentum of electron in 4th orbit of hydrogen atom is

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

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

C
$$\dfrac { nh }{ 2\pi } $$

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

Solution

The angular momentum of electron in $$4th$$ orbit of hydrogen atom is $$\dfrac{{nh}}{{2\pi }}$$
Hence, Option $$C$$ is correct.
Question 4
1 / -0

In the Geiger - Marsden scattering experiment, find the distance of closest approach to the nucleus of a $$7.7Mev$$ $$\alpha$$ - particle before it comes momentaily to rest and reverses its direction. ($$'Z'$$ for gold nucleus $$= 79$$)

A
$$10fm$$

B
$$20fm$$

C
$$30fm$$

D
$$40fm$$
Question 5
1 / -0

According to Bohr's model, if the kinetic energy of an electron in $$2^{nd}$$ orbit of $$He^{+}$$ is $$x$$, then what should be the ionisaion energy of the electron revolving in $$3^{rd}$$ orbit of $$m^{5+} ions>$$

A
$$X$$

B
$$4X$$

C
$$x/4$$

D
$$2X$$

Solution

$$\begin{array}{l} X=13\cdot 6\frac { { { { \left( 2 \right) }^{ 2 } } } }{ { { { \left( 2 \right) }^{ 2 } } } } \\ =13\cdot 6 \\ X'=13\cdot \frac { { { { \left( 6 \right) }^{ 2 } } } }{ { { { \left( 3 \right) }^{ 2 } } } } \\ =13\cdot 6\times \frac { { 36 } }{ 9 } \\ =4X \\ Hence,\, option\, B\, is\, correct\, answer. \end{array}$$
Question 6
1 / -0

Energy of an electron is given by $$E=-2.178\times { 10 }^{ -18 }J\left( \dfrac { { Z }^{ 2 } }{ { n }^{ 2 } } \right) $$ Wavelength of light required to excite an electron in an hydrogen atom from level n=1 to n=2 will be:-`
$$\left( h=6.62\times { 10 }^{ -34 }Js\quad and\quad c=3.0\times { 10 }^{ 8 }{ ms }^{ -1 } \right) $$

A
$$1.214\times { 10 }^{ -7 }m$$

B
$$2.816\times { 10 }^{ -7 }m$$

C
$$6.500\times { 10 }^{ -7 }m$$

D
$$8.500\times { 10 }^{ -7 }m$$

Solution

$$\begin{array}{l} \dfrac { 1 }{ \lambda } ={ R_{ H } }{ Z^{ 2 } }\left( { \dfrac { 1 }{ { n_{ 1 }^{ 2 } } } -\dfrac { 1 }{ { n_{ 2 }^{ 2 } } } } \right) \\ Z=1 \\ { n_{ 1 } }=1 \\ { n_{ 2 } }=2 \\ \lambda =1.24\times { 10^{ -7 } }m \end{array}$$
Question 7
1 / -0

The difference between the wave number of $${ 1 }^{ st }$$ line of Balmer series of paschen series for $${ Li }^{ 2+ }ion$$ is:

A
$$\dfrac { R }{ 36 } $$

B
$$\dfrac { 5R }{ 36 } $$

C
$$4R$$

D
$$\dfrac { R }{ 4 } $$

Solution

$$\begin{array}{l} \frac { 1 }{ \lambda } =R{ t^{ 2 } }\left[ { \frac { 1 }{ { n_{ 1 }^{ 2 } } } -\frac { 1 }{ { n_{ 2 }^{ 2 } } } } \right] \\ 1st\, \, line\, \, of\, \, polymer\, \, series=\alpha \to 3 \\ Last\, \, \, n\, \, of\, \, polymer\, \, series\, \, =3\to \infty \\ L_{ 1 }^{ st }=t=3 \\ { \overline { V } _{ B } }-{ \overline { V } _{ L } }=R{ \left( 3 \right) ^{ 2 } }\left[ { \frac { 1 }{ { { \alpha ^{ 2 } } } } -\frac { 1 }{ { { 3^{ 2 } } } } } \right] -R{ \left( 3 \right) ^{ 2 } }\left[ { \frac { 1 }{ { { 3^{ 2 } } } } -\frac { 1 }{ \infty } } \right] \\ =R{ \left( 3 \right) ^{ 2 } }\left[ { \frac { 1 }{ 4 } -\frac { 1 }{ 9 } -\frac { 1 }{ 9 } } \right] \\ =R\times 9\left[ { \frac { 1 }{ 4 } -\frac { 2 }{ 9 } } \right] \\ =R\times 9\times \frac { { 9-8 } }{ { 36 } } \\ { \overline { V } _{ B } }-{ \overline { V } _{ L } }=R\times 9\times \frac { 1 }{ { 26 } } \\ The\, \, difference=\frac { R }{ 4 } \end{array}$$
Hence, Option $$D$$ is correct.
Question 8
1 / -0

Two electron separated by a distance $$r$$ experiences force $$F$$ between them . Then force between a proton and a singly ionized helium atom separated by a distance $$2r$$ is

A
$$4F$$

B
$$2F$$

C
$$\frac { F } {2}$$

D
$$\frac { F } {4 }$$

Solution

According to Coulomb's law
$$\begin{array}{l} F=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\dfrac { { \left( e \right) \left( e \right) } }{ { { r^{ 2 } } } } \\ { F_{ 1 } }=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\dfrac { { \left( e \right) \left( e \right) } }{ { { r^{ 2 } } } } \end{array}$$
Charge on proton is equal to charge on electron and charge on singly Ionized
atom separated by distance $$'2r'$$
$$\begin{array}{l} { F_{ 2 } }=\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\frac { { \left( { e\lambda c } \right) } }{ { { { \left( { 2r } \right) }^{ 2 } } } } \\ =\dfrac { 1 }{ { 4\pi { E_{ 0 } } } } .\dfrac { { { e^{ 2 } } } }{ { 4{ r^{ 2 } } } } \\ \dfrac { { { F_{ 1 } } } }{ { { F_{ 2 } } } } =\dfrac { F }{ 4 } \end{array}$$
Option D.
Question 9
1 / -0

Hydrogen atom in ground state is excited by a monochromatic radiation of $$\lambda =975\mathring { A } $$ Number of spectral lines in the resulting spectrum emitted will be

A
$$3$$

B
$$2$$

C
$$6$$

D
$$10$$
Question 10
1 / -0

The ratio of the frequencies of the long wavelength limits of the Lyman and Balmer series of hydrogen is:

A
$$27 : 5$$

B
$$5 : 27$$

C
$$4 : 1$$

D
$$1 : 4$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Atoms Test - 57

Result

Atoms Test - 57

Score

-

Rank

-

SHARING IS CARING

Question 1

$$2.645\ A^o$$

$$1.322\ A^o$$

$$0.2645\ A^{o}$$

$$None\ of\ these$$

Solution

Question 2

$$1:\dfrac { 1 }{ 2 } :\dfrac { 1 }{ 3 } $$

$$1:2:3$$

$$1:4:9$$

$$1:8:27$$

Solution

Question 3

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

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

$$\dfrac { nh }{ 2\pi } $$

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

Solution

Question 4

$$10fm$$

$$20fm$$

$$30fm$$

$$40fm$$

Question 5

$$X$$

$$4X$$

$$x/4$$

$$2X$$

Solution

Question 6

$$1.214\times { 10 }^{ -7 }m$$

$$2.816\times { 10 }^{ -7 }m$$

$$6.500\times { 10 }^{ -7 }m$$

$$8.500\times { 10 }^{ -7 }m$$

Solution

Question 7

$$\dfrac { R }{ 36 } $$

$$\dfrac { 5R }{ 36 } $$

$$4R$$

$$\dfrac { R }{ 4 } $$

Solution

Question 8

$$4F$$

$$2F$$

$$\frac { F } {2}$$

$$\frac { F } {4 }$$

Solution

Question 9

$$3$$

$$2$$

$$6$$

$$10$$

Question 10

$$27 : 5$$

$$5 : 27$$

$$4 : 1$$

$$1 : 4$$

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.