Atoms Test

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Atoms Test - 69

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Weekly Quiz Competition

Question 1
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Directions For Questions

The figure shows an energy level diagram of the hydrogen atom. Several transitions are marked as $$I, II, III, ...$$. The diagram is only indicative and not to scale.

...view full instructions

In which transition is a Balmer series photon absorbed?

A
$$II$$

B
$$III$$

C
$$IV$$

D
$$VI$$

Solution

Absorption occurs when electron excites to higher level.
For Balmer, electron has to excite from $$n=2$$.
Question 2
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The ratio between Bohr radii is

A
1 : 2 : 3

B
2 : 4 : 6

C
1 : 4 : 9

D
1 : 3 : 5

Solution

Radius $$\propto n^2$$
Hence, $$R_1:R_2:R_3 = 1:4:9$$
Question 3
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If the atom $$_{100} Fm^{257}$$ follows the Bohr model and the radius of $$_{100} Fm^{257}$$ is n times the Bohr radius, then find $$n$$.

A
$$100$$

B
$$200$$

C
$$4$$

D
$$1/4$$

Solution

For an atom following Bohr's model, the radius is given by $$\displaystyle r_m=\frac{r_0m^2}{Z}$$
where $$r_0=$$Bohr's radius and $$m=$$ orbit number.
For $$Fm, m=5$$ (Fifth orbit in which the outermost electron is present)
$$\displaystyle\therefore r_m=\frac{r_05^2}{100}=nr_0$$ (given) $$\displaystyle\Rightarrow n=\frac{1}{4}$$
Question 4
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The radius of the first orbit of hydrogen is $$r_H$$, and the energy in the ground state is $$-13.6$$eV. Considering a $$\mu^-$$-particle with a mass $$207$$ $$m_e$$ revolving round a proton as in Hydrogen atom, the energy and radius of proton and $$\mu^-$$ combination respectively in the first orbit are:
[Assume nucleus to be stationary]

A
$$-13.6\times 207\ eV$$, $$\displaystyle\frac{r_H}{207}$$

B
$$-207\times 13.6\ eV$$, $$207$$ $$r_H$$

C
$$\displaystyle\frac{-13.6}{207}\ eV$$, $$\displaystyle\frac{r_H}{207}$$

D
$$\displaystyle\frac{-13.6}{207}\ eV$$, $$207$$ $$r_H$$

Solution

Radius of ground state, $$r = \dfrac{h^2}{4\pi ^2 mKe^2} $$
$$\implies$$ $$r \propto \dfrac{1}{m}$$

$$\implies$$ $$\dfrac{r_{\mu}}{r_H} = \dfrac{m_e}{m_{\mu}} = \dfrac{m_e}{207 m_e}$$

$$\implies$$ $$r_{\mu } = \dfrac{r_H}{207}$$

Energy of ground state, $$E = - \dfrac{Ke^2 }{2 r}$$
$$\implies$$ $$E \propto m$$ $$(\because r \propto \dfrac{1}{m})$$

$$\implies$$ $$\dfrac{E_{\mu}}{ -13.6} = \dfrac{m_{\mu}}{m_e} = \dfrac{207m_e}{ m_e}$$

$$\implies$$ $$E_{\mu} = -13.6 \times 207$$ $$eV$$
Question 5
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A donor atom in a semiconductor has a loosely bound electron. The orbit of this electron is considerably affected by the semiconductor material but behaves in many ways like an electron orbiting a hydrogen nucleus. Given that the electrons has an effective mass of $$0.07\ m_e$$ (where $$m_{e}$$ is mass of the free electron) and the space in which it moves has a permittivity $$13\epsilon_{0}$$, then the radius of the electron's lowermost energy orbit will be close to (The Bohr radius of the hydrogen atom is $$0.53\overset {\circ}{A})$$.

A
$$0.53\overset {\circ}{A}$$

B
$$243\overset {\circ}{A}$$

C
$$10\overset {\circ}{A}$$

D
$$100\overset {\circ}{A}$$

Solution

$$\dfrac {mv^{2}}{r} = \dfrac {KQ^{2}}{r^{2}}$$
and
$$mvr = \dfrac {nh}{2\pi}\implies mv = \dfrac{nh}{2 \pi r}$$

and $$ \dfrac{m^2v^2}{mr}=\dfrac{KQ^2}{r^2}$$
$$ \dfrac{n^2h^2}{4 \pi^2mr^3}=\dfrac{1}{4\pi \epsilon_0}\dfrac{Q^2}{r^2}$$
$$ \dfrac{n^2h^2}{4\pi^2(0.07m_e)r}=\dfrac{1}{4\pi(13) \epsilon_0} Q^2$$
$$ r=\dfrac{13n^2h^2}{4\pi^2kQ^2(0.07)m_e}$$
$$r=\dfrac{13}{0.07} \times r_0$$ where $$r_0=bohr \ radius$$
$$r=\dfrac{13}{0.07} \times 0.53 \overset{\circ}{A}$$
$$r \approx 100\overset {\circ}{A}$$.
Question 6
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The number of revolutions per second made by an electron in the first Bohr's orbit of hydrogen atom is(Given, $$r=0.53\overset{o}{A}$$).

A
$$6\times 10^5$$Hz

B
$$2.5\times 10^{12}$$Hz

C
$$1.9\times 10^{13}$$Hz

D
$$6.5\times 10^{15}$$Hz

Solution

From Bohr's theory
$$mvr=\displaystyle\frac{nh}{2\pi}$$
For first orbit $$n=1$$
$$m(r\omega )r=\displaystyle\frac{1\times h}{2\pi}$$
$$\Rightarrow mr^22\pi f=\displaystyle\frac{h}{2\pi}$$
$$\Rightarrow \displaystyle f=\frac{h}{4\pi^2mv^2}$$
$$=\displaystyle\frac{6.6\times 10^{-34}}{4\times (3.14)^2\times 9.1\times 10^{-31}\times (0.33\times 10^{-10})^2}$$.
=$$6.5\times10^{15} Hz$$
Question 7
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To calculate the size of a hydrogen anion using the Bohr model, we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum $$\hbar = h/2\pi$$, and taking electron interaction into account the radius of the orbit in terms of the Bohr radius of hydrogen atom $$a_{B}= \dfrac {4\pi \epsilon_{0}\hbar^{2}}{me^{2}}$$ is

A
$$a_{B}$$

B
$$\dfrac {4}{3}a_{B}$$

C
$$\dfrac {2}{3}a_{B}$$

D
$$\dfrac {3}{2}a_{B}$$

Solution

Let the velocity of the electron be $$v$$.
Angular momentum of electron $$mvr = \hbar$$ (given)
Net electrostatic force acting on the electron $$F_e = \dfrac{e^2}{4\pi\epsilon_o r^2} - \dfrac{e^2}{4\pi\epsilon_o (2r^2)} = \dfrac{3}{4}\dfrac{e^2}{4\pi\epsilon_o r^2}$$
Centrifugal force $$F_c = \dfrac{mv^2}{r} = \dfrac{(mvr)^2}{r\times mr^2} = \dfrac{\hbar^2}{mr^3}$$
But $$F_c = F_e$$
$$\therefore$$ $$\dfrac{\hbar^2}{mr^3} = \dfrac{3}{4}\dfrac{e^2}{4\pi\epsilon_o r^2}$$
$$\implies \ r = \dfrac{4}{3}\dfrac{4\pi\epsilon_o \hbar^2}{me^2} = \dfrac{4}{3}a_B$$
Question 8
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In the Bohr model of a hydrogen atom, the centripetal force is furnished by the coulomb attraction between the proton and the electron. If $$a_0$$ is the radius of the ground state orbit, m is the mass and e is the charge on the election and $$\varepsilon_0$$ is the vacuum permittivity, the speed of the electron is?

A
$$0$$

B
$$\displaystyle\frac{e}{\sqrt{\varepsilon_0a_0m}}$$

C
$$\displaystyle\frac{e}{\sqrt{4\pi\varepsilon_0a_0m}}$$

D
$$\sqrt{\displaystyle\frac{4\pi\varepsilon_0a_0m}{e}}$$

Solution

Centripetal force$$=\dfrac { m{ v }^{ 2 } }{ { a }_{ \circ } } \rightarrow (1)$$
Colounbic force$$=\dfrac { 1 }{ 4\pi { \varepsilon }_{ \circ } } \dfrac { e.e }{ { a }^{ 2 } } \rightarrow (2)$$
$$\therefore$$ Centripetal force= Coloumbic force
$$\rightarrow \dfrac { m{ v }^{ 2 } }{ { a }_{ \circ } } \dfrac { 1 }{ 4\pi { \varepsilon }_{ \circ } } \dfrac { { e }^{ 2 } }{ { { a }_{ \circ } }^{ 2 } } \\ \rightarrow { v }^{ 2 }=\dfrac { { e }^{ 2 } }{ 4\pi { \varepsilon }_{ \circ }{ a }_{ \circ }m } \\ or\quad v=\dfrac { e }{ \sqrt { 4\pi { \varepsilon }_{ \circ }{ a }_{ \circ }m } } $$
Question 9
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To calculate the size of a hydrogen anion using the Bohr model, we assume that its two electrons move in an orbit such that they are always on diametrically opposite sides of the nucleus. With each electron having the angular momentum equal to $$\dfrac{h}{ 2 \pi}$$, and taking electron interaction into account the radius of the orbit in terms of the Bohr radius of hydrogen atom $$a_B=\displaystyle\frac{4\pi \epsilon_0h^2}{me^2}$$ is?

A
$$a_B$$

B
$$\displaystyle\frac{4}{3}a_B$$

C
$$\displaystyle\frac{2}{3}a_B$$

D
$$\displaystyle\frac{3}{2}a_B$$

Solution
Question 10
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Given mass number of glod=197, density of gold=19.7 g per $$cm^3$$, Avogadro's number = $$6\times 10^{23}$$.The radius of the gold atom is approximately:

A
$$1.5 \times 10^{-8}m$$

B
$$1.5 \times 10^{-9}m$$

C
$$1.5 \times 10^{-10}m$$

D
$$1.5 \times 10^{-12}m$$

Solution

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Re-Attempt Weekly Quiz Competition

Atoms Test - 69

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Atoms Test - 69

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SHARING IS CARING

Question 1

$$II$$

$$III$$

$$IV$$

$$VI$$

Solution

Question 2

1 : 2 : 3

2 : 4 : 6

1 : 4 : 9

1 : 3 : 5

Solution

Question 3

$$100$$

$$200$$

$$4$$

$$1/4$$

Solution

Question 4

$$-13.6\times 207\ eV$$, $$\displaystyle\frac{r_H}{207}$$

$$-207\times 13.6\ eV$$, $$207$$ $$r_H$$

$$\displaystyle\frac{-13.6}{207}\ eV$$, $$\displaystyle\frac{r_H}{207}$$

$$\displaystyle\frac{-13.6}{207}\ eV$$, $$207$$ $$r_H$$

Solution

Question 5

$$0.53\overset {\circ}{A}$$

$$243\overset {\circ}{A}$$

$$10\overset {\circ}{A}$$

$$100\overset {\circ}{A}$$

Solution

Question 6

$$6\times 10^5$$Hz

$$2.5\times 10^{12}$$Hz

$$1.9\times 10^{13}$$Hz

$$6.5\times 10^{15}$$Hz

Solution

Question 7

$$a_{B}$$

$$\dfrac {4}{3}a_{B}$$

$$\dfrac {2}{3}a_{B}$$

$$\dfrac {3}{2}a_{B}$$

Solution

Question 8

$$0$$

$$\displaystyle\frac{e}{\sqrt{\varepsilon_0a_0m}}$$

$$\displaystyle\frac{e}{\sqrt{4\pi\varepsilon_0a_0m}}$$

$$\sqrt{\displaystyle\frac{4\pi\varepsilon_0a_0m}{e}}$$

Solution

Question 9

$$a_B$$

$$\displaystyle\frac{4}{3}a_B$$

$$\displaystyle\frac{2}{3}a_B$$

$$\displaystyle\frac{3}{2}a_B$$

Solution

Question 10

$$1.5 \times 10^{-8}m$$

$$1.5 \times 10^{-9}m$$

$$1.5 \times 10^{-10}m$$

$$1.5 \times 10^{-12}m$$

Solution

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