Atoms Test

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Atoms Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

The figure shows the velocity and acceleration of a point like body at the initial moment of its motion. The acceleration vector of the body remains constant. The minimum radius of curvature of trajectory of the body is (in m)

A
2 m

B
4 m

C
8 m

D
16 m
Question 2
1 / -0

In a hydrogen atom, electron revolves in a circular orbit of radius $$0.53\overset {\circ}{A}$$ with a velocity of $$2.2\times 10^{6} ms^{-1}$$. If the mass of electron is $$9.0\times 10^{-31} kg$$, find its angular momentum.

A
$$05\times 10^{-34} kg\ m^{2}s^{-1}$$.

B
$$1.05\times 10^{-34} kg\ m^{2}s^{-1}$$.

C
$$2.05\times 10^{-34} kg\ m^{2}s^{-1}$$.

D
$$3.05\times 10^{-34} kg\ m^{2}s^{-1}$$.

Solution

$$\text{Given- $*$ Radius $r=0.53 A^{\circ} *$ velocity $v=2.2 \times 10^{6} \mathrm{~m} /s$}$$
$$ *\text { mass } m=9 \times 10^{-31} \mathrm{~kg}$$
$$\text{Orbital an gular momentum}$$ =m v r
$$\begin{aligned}m v r &=9 \times 10^{-31} \mathrm{~kg} \times 2.2\times 10^{6} \mathrm{~m} /s \times 0.53 \times 10^{-10}\mathrm{~m}\\&=1.05 \times 10^{-34}\mathrm{~kg}\mathrm{m}^{2}/\mathrm{~s}\end{aligned}$$
Question 3
1 / -0

According to Bohr's theory of hydrogen atom, the speed of the electron, its energy and the radius of its orbit varies with the principal quantum number $$n$$, respectively as:

A
$$\cfrac { 1 }{ n } ,\cfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }\quad $$

B
$$\cfrac { 1 }{ n } ,{ n }^{ 2 },\cfrac { 1 }{ { n }^{ 2 } } $$

C
$${ n }^{ 2 },\cfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }$$

D
$$n,\cfrac { 1 }{ { n }^{ 2 } } ,\cfrac { 1 }{ { n }^{ 2 } } $$

Solution
Question 4
1 / -0

Taking the Bohr radius as $$a_{0}=35pm$$, the radius of $$Li^{++}$$ion in its ground state, on the basis of Bohr's model, will be about

A
$$53pm$$

B
$$27\ pm$$

C
$$18\ pm$$

D
$$13\ pm$$

Solution

$$\begin{array}{l}\text { Given, } \\\text { Bohr's radius }=53 \mathrm{pm} \\\text { we know, } \\\text { Bohr's radius }=\frac{0.529 n^{2}}{2}\end{array}$$
$$\begin{array}{l}\text { if } n=1 & Z=1 \text { , the radius obtained is bohr's radius} \\\end{array}$$
$$\begin{array}{l}\text { radius }\left(a_{0}\right)=\frac{0.529(1)^{2}}{1} A=53\mathrm{Pm} \\\text { for } Li^{++} \quad n=1 \quad(\text { ground state }) \\\text { and } z=3\end{array}$$
$$\begin{aligned}\text { radius } &=\frac{53\cdot n^{2}}{2} p_{m}=\frac{53(1)^{2}}{3} \\r &=17.6 \approx 18 \mathrm{pm} \\\text { hence option (c) is wrrect. }\end{aligned}$$
Question 5
1 / -0

An ionic atom equivalent to hydrogen atom has wavelength equal to $$\dfrac{1}{4}$$ of the wavelengths of hydrogen lines. The ion will be

A
$$He^+$$

B
$$Li^{++}$$

C
$$Ne^{9+}$$

D
$$Na^{10+}$$

Solution

$$\text{By Rydberg's formula ,we have -}$$
$$\begin{array}{l}R_{H}=\text { Rydberg constant } \\\lambda=\text { wavelength } \\Z=\text { atomic num. }\end{array}$$

$$\begin{array}{l}\text { * For a particular member of spectral line }\\\text { we have - } \\\qquad \begin{aligned}\frac{1}{\lambda} &\alpha z^{2} \quad \Rightarrow\left[\lambda \alpha \frac{1}{z^{2}}\right] \\& \Rightarrow \lambda=\frac{k}{z^{2}} \text { (let) }-(1) \\\Rightarrow & \lambda_{H}=k\end{aligned}\end{array}$$

$$\begin{array}{l}\text { For any other ionic atom for which } \\\text { wavelength }=\frac{\lambda_{H}}{4}=\frac{k}{4}-(2) \\\text { from eq (1) and (2) } \\\text { we get } \frac{k}{z^{2}}=\frac{k}{4}\\\Rightarrow z^{2}=4 \\\Rightarrow z=2\quad\left(\mathrm{He}^{+}\right) \text {atom. } \\\text { Hence, option (A) is correct. }\end{array}$$
Question 6
1 / -0

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of $$108.5nm$$. the ground state energy of an electron of this ion will be

A
$$3.4eV$$

B
$$13.6eV$$

C
$$54.4eV$$

D
$$122.4eV$$

Solution

For third of Balmer series $$n_1=2,\ n_2=5$$

$$\therefore \dfrac{1}{\lambda}=RZ^2 \left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2} \right]$$

gives $$Z^2=\dfrac{n_1^2 n_2^2}{(n_2^2-n_1^2) \lambda R}$$

On putting values $$Z=2$$

From $$E=-\dfrac{13.6Z^2}{n^2}=\dfrac{-13.6(2)^2}{(1)^2}=-54.4\ eV$$
Question 7
1 / -0

The electron in a hydrogen atom makes a transition $${n}_{1}\rightarrow {n}_{2}$$ whose $${n}_{1}$$ and $${n}_{2}$$ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is $$1/27$$ of that in the final state. The possible values of $${n}_{1}$$ and $${n}_{2}$$ are

A
$${n}_{1}=4,{n}_{2}=2$$

B
$${n}_{1}=3,{n}_{2}=1$$

C
$${n}_{1}=8,{n}_{2}=1$$

D
$${n}_{1}=6,{n}_{2}=3$$

Solution
Question 8
1 / -0

If the kinetic energy of an electron in the first orbit of $$H$$ atom is $$13.6 eV$$. Then the total energy of an electron in the second orbit of $${He}^{+}$$ is:

A
$$13.6 eV$$

B
$$3.4 eV$$

C
$$-13.6 eV$$

D
$$-3.4 eV$$

Solution

$$\begin{array}{l}\text { Given- * kinetic energy (KE) of e" in first orbit of } \\\text { H-atom is } 13.6 \mathrm{eV.} \\\text { we know that - } \\\qquad k E=13.6 \mathrm{z}^{2}/n^2 \quad n \rightarrow n^{\text {th }} \text { orbit } \\\quad z \rightarrow \text { atomic number. }\end{array}$$

$$\begin{array}{l}\text { we also know that - } \\\text { (Total Energy) }=-(\text { kinetic Energy) } \\\Rightarrow \quad T E=-K E\\\Rightarrow T E=-13.6 \frac{z^{2}}{n^{2}} e V \\\text { Now, for } H e^{+} \text {atom }(z=2) \text { in second } \text { orbit }(n=2)\\\qquad T E=-13.6 \text { eV } \frac{4}{4}=-13.6\mathrm{eV}\end{array}$$
Question 9
1 / -0

Which of the following weighs the least?

A
2 g atom of N (atomic wt of N =14)

B
$$ 3 \times 10^{23} $$ atoms of C (atomic wt. of C=12)

C
1 mole of S ( atomic weight of S = 32 )

D
7 g silver ( atomic wt. of Ag =108)

Solution
Question 10
1 / -0

The first line of Balmer series has wavelength $$6563\mathring {A}$$. what will be the wavelength of the first memeber of lymais series.

A
$$1215\mathring{A}$$

B
$$2500\mathring {A}$$

C
$$7500\mathring {A}$$

D
$$600\mathring {A}$$

Solution

$$\begin{array}{l}\text { Given, } \\\text { First line of Balmer series has wavelength }=6563 \hat{A}\end{array}$$
$$\begin{array}{l}\text { We know, } \\\text { for Balmer series } \\\text { line is from } n_{1}=2 \text { to } n_{2}=3,4,5\ldots\end{array}$$
$$\begin{array}{l}\text { and for laymen series } \\\text { line is from } n_{1}=1 \text { to } n_{2}=2,3,4 \ldots\end{array}$$
$$\begin{array}{l}\text { So, First line of balmer series mean } \\\qquad n_{1}=2 \text { to } n_{2}=3\end{array}$$
$$\text { we know }$$
$$\frac{1}{\lambda}=R z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$$
$$R z^{2}=\frac{36}{6563 \times 5} \ldots \text { (1) }$$
$$\begin{array}{l}\text { for first line of lymen series } \\\qquad n=1 \text { to } n=2\end{array}$$
$$\begin{aligned}\frac{1}{\lambda} &=R z^{2}\left(\frac{1}{(1)^{2}}-\frac{1}{4}\right) \\\frac{1}{\lambda} &=\frac{36}{6563 \times 5}\left(\frac{3}{4}\right) \\\lambda &=\frac{6563 \times 5 \times 4}{36 \times 3} \\\lambda &=1215.3 \dot A\end{aligned}$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 72

Result

Atoms Test - 72

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SHARING IS CARING

Question 1

2 m

4 m

8 m

16 m

Question 2

$$05\times 10^{-34} kg\ m^{2}s^{-1}$$.

$$1.05\times 10^{-34} kg\ m^{2}s^{-1}$$.

$$2.05\times 10^{-34} kg\ m^{2}s^{-1}$$.

$$3.05\times 10^{-34} kg\ m^{2}s^{-1}$$.

Solution

Question 3

$$\cfrac { 1 }{ n } ,\cfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }\quad $$

$$\cfrac { 1 }{ n } ,{ n }^{ 2 },\cfrac { 1 }{ { n }^{ 2 } } $$

$${ n }^{ 2 },\cfrac { 1 }{ { n }^{ 2 } } ,{ n }^{ 2 }$$

$$n,\cfrac { 1 }{ { n }^{ 2 } } ,\cfrac { 1 }{ { n }^{ 2 } } $$

Solution

Question 4

$$53pm$$

$$27\ pm$$

$$18\ pm$$

$$13\ pm$$

Solution

Question 5

$$He^+$$

$$Li^{++}$$

$$Ne^{9+}$$

$$Na^{10+}$$

Solution

Question 6

$$3.4eV$$

$$13.6eV$$

$$54.4eV$$

$$122.4eV$$

Solution

Question 7

$${n}_{1}=4,{n}_{2}=2$$

$${n}_{1}=3,{n}_{2}=1$$

$${n}_{1}=8,{n}_{2}=1$$

$${n}_{1}=6,{n}_{2}=3$$

Solution

Question 8

$$13.6 eV$$

$$3.4 eV$$

$$-13.6 eV$$

$$-3.4 eV$$

Solution

Question 9

2 g atom of N (atomic wt of N =14)

$$ 3 \times 10^{23} $$ atoms of C (atomic wt. of C=12)

1 mole of S ( atomic weight of S = 32 )

7 g silver ( atomic wt. of Ag =108)

Solution

Question 10

$$1215\mathring{A}$$

$$2500\mathring {A}$$

$$7500\mathring {A}$$

$$600\mathring {A}$$

Solution

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