Atoms Test

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Atoms Test - 11

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Weekly Quiz Competition

Question 1
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Directions For Questions

The key feature of Bohrs theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohrs quantization condition.

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A diatomic molecule has moment of inertia $$I$$. By Bohrs quantization condition its rotational energy in the $$n^{th}$$ level ($$n = 0$$ is not allowed) is

A
$$\displaystyle \frac{1}{\mathrm{n}^{2}}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

B
$$\displaystyle \frac{1}{\mathrm{n}}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

C
$$\displaystyle \mathrm{n}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

D
$$\displaystyle \mathrm{n}^{2}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

Solution

$$\displaystyle \mathrm{L}=\frac{\mathrm{n}\mathrm{h}}{2\pi}$$
$$\mathrm{K}.\mathrm{E}.\ =\displaystyle \frac{\mathrm{L}^{2}}{2\mathrm{I}}=(\frac{\mathrm{n}\mathrm{h}}{2\pi})^{2}\frac{1}{2\mathrm{I}}$$
Question 2
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Directions For Questions

The key feature of Bohrs theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohrs quantization condition.

...view full instructions

In a $$\mathrm{C}\mathrm{O}$$ molecule, the distance between $$\mathrm{C}$$ (mass $$=12$$ a.m.u) and $$\mathrm{O}$$ (mass $$=16$$ a.m.u.), where 1 a.m.u $$=\displaystyle \frac{5}{3}\times 10^{-27} kg,$$ is close to

A
$$2.4\times 10^{-10}\mathrm{m}$$

B
$$1.9\times 10^{-10}\mathrm{m}$$

C
$$1.3\times 10^{-10}\mathrm{m}$$

D
$$4.4\times 10^{-11}\mathrm{m}$$

Solution

$$\displaystyle \mathrm{r}_{1}=\frac{\mathrm{m}_{2}\mathrm{d}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$$ and $$\displaystyle \mathrm{r}_{2}=\frac{\mathrm{m}_{1}\mathrm{d}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$$

$$\mathrm{I}=\mathrm{m}_{1}\mathrm{r}_{1}^{2}+\mathrm{m}_{2}\mathrm{r}_{2}^{2}$$

$$\therefore \mathrm{d}=1.3\times 10^{-10}\mathrm{m}$$.
Question 3
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The electrostatic energy of $$Z$$ protons uniformly distributed throughout a spherical nucleus of radius $$R$$ is given by
$$\displaystyle E=\frac { 3 }{ 5 } \frac { Z\left( Z-1 \right) { e }^{ 2 } }{ 4\pi { \varepsilon }_{ 0 }R } $$
The measured masses of the neutron, $$\displaystyle _{ 1 }^{ 1 }{ H },\, _{ 7 }^{ 15 }{ N }$$ and $$\displaystyle _{ 8 }^{ 15 }{ O }$$ are $$1.008665\ u, 1.007825 \ u, 15.000109\ u$$ and $$15.003065\ u$$, respectively. Given that the radii of both the $$\displaystyle _{ 7 }^{ 15 }{ N }$$ and $$\displaystyle _{ 8 }^{ 15 }{ O }$$ nuclei are same, $$\displaystyle 1u=931.5\quad { MeV }/{ { c }^{ 2 } }$$ (c is the speed of light) and $$\displaystyle { { e }^{ 2 } }/{ \left( 4\pi { \varepsilon }_{ 0 } \right) }=1.44\,MeV\,fm$$. Assuming that the difference between the binding energies of $$\displaystyle _{ 7 }^{ 15 }{ N }$$ and $$\displaystyle _{ 8 }^{ 15 }{ O }$$ is purely due to the electrostatic energy, the radius of either of the nuclei is $$(1\, fm = 10^{-15}m)$$:

A
$$2.85\ fm$$

B
$$3.03\ fm$$

C
$$3.42\ fm$$

D
$$3.80\ fm$$

Solution

For $$^{15}_{7} N$$ : $$Z_1 = 7$$ $$N_1 = 8$$
For $$^{15}_{8} O$$ : $$Z_2 = 8$$ $$N_2 = 7$$

Difference in the electrostatic energy, $$\Delta E = E_2 -E_1 $$
$$\Delta E = \dfrac{3}{5} \bigg(\dfrac{8 (8-1)}{R} - \dfrac{7(7-1)}{R}\bigg) \times 1.44 = \dfrac{12.096}{R}$$ $$MeV $$ $$fm$$

Mass defect, $$\Delta M = Z M_H + N M_n - M_{atom}$$
$$\therefore$$ For $$^{15}_{7} N$$    $$\Delta M_1 = 7(1.007825) + 8(1.008665) - 15.000109 = 0.12399$$ u

For $$^{15}_{8} O$$,   $$\Delta M_2 = 8(1.007825) + 7(1.008665) - 15.003065 = 0.12019$$ u

Difference in binding energy, $$\Delta B.E = [\Delta M_1 - \Delta M_2] \times 931.5$$ $$MeV$$
$$\implies$$ $$\Delta B.E = [0.12399 -0.12019] \times 931.5 = 3.54$$ $$MeV$$

According to the question, $$\dfrac{12.096}{R} = 3.54$$ $$\implies R = 3.42$$ $$fm$$
Question 4
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The electric potential between a proton and an electron is given by $$V=\displaystyle \mathrm{V}_{0}\ln\frac{\mathrm{r}}{\mathrm{r}_{0}}$$ , where $$\mathrm{r}_{0}$$ is a constant. Assuming Bohr's model to be applicable, write variation of $$\mathrm{r}_{\mathrm{n}}$$ with $$\mathrm{n},\ \mathrm{n}$$ being the principal quantum number?

A
$$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}$$

B
$$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}$$

C
$$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}^{2}$$

D
$$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}^{2}$$

Solution

Electric potential between proton and electron in nth orbit is given as: $$V = V_o ln\dfrac{r_n}{r_o}$$
Thus the coulomb force $$|F_c| = e\dfrac{dV}{dr} = e\dfrac{V_o}{r_n}$$
This coulombic force is balance by the centripetal force.
$$\therefore \dfrac{mv^2}{r_n} = e\dfrac{V_o}{r_n} \implies v = \sqrt{\dfrac{e V_o}{m}}$$ means that speed is constant
Now from $$m v r_n = \dfrac{nh}{2\pi} \implies r_n \propto n$$ (as $$v = constant$$)
Question 5
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For which one of the following, Bohr model is not valid?

A
Singly ionised helium atom $$(He^+)$$

B
Deuteron atom

C
Singly ionised neon atom $$(Ne^+)$$

D
Hydrogen atom

Solution

Bohr's model is valid only for the single electronic systems. Therefore, it will not be valid for a singly ionised neon atom as it has more electrons.
Question 6
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When an $$\alpha$$-particle of mass $$'m'$$ moving with velocity $$'v'$$ bombards on a heavy nucleus of charge $$'Ze'$$, its distance of closest approach from the nucleus depends on $$m$$ as:

A
$$\dfrac { 1 }{ m } $$

B
$$\dfrac { 1 }{ \sqrt { m } } $$

C
$$\dfrac { 1 }{ { m }^{ 2 } } $$

D
$$m$$

Solution

At closest distance of approach, the kinetic energy of the particle will completely convert into potential energy.

$$\Rightarrow \dfrac{1}{2}mv^2=\dfrac{KQq}{d}$$

$$\Rightarrow d\propto \dfrac{1}{m}$$
Question 7
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The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong?

A
$$\gamma$$ rays

B
x rays

C
Infra - red rays

D
Ultraviolet rays

Solution

$$\lambda = \dfrac{hc}{E} =\dfrac{1240 ev nm}{15\times 10^3} =0.083 \ nm $$

Since the order of wavelength is between $$10\ nm$$ and $$.001\ nm$$, it belongs to X-ray. part of the spectrum.
Question 8
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The value of Planck's constant is $$6.63\times 10^{-34}Js$$. The speed of light $$3\times 10^{17}\:nm\:s^{-1}$$. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of $$6\times 10^{15}s^
{-1}$$?

A
$$10$$

B
$$25$$

C
$$50$$

D
$$75$$

Solution

$$\displaystyle v=\frac{c}{\lambda }$$
$$\therefore \displaystyle \lambda =\frac{3\times 10^{17}nms^{-1}}{6\times 10^{15}s^{-1}}=50\:nm$$
Question 9
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The transition from the state $$n=3$$ to $$n=1$$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from-

A
$$3\rightarrow 2$$

B
$$4\rightarrow 2$$

C
$$4\rightarrow 3$$

D
$$2\rightarrow 1$$

Solution

Infrared radiation found in Paschen, Bracket and Fund series and it is obtained when $$"{e}^{-}"$$ transition is from a high energy level to minimum $$3$$rd energy level.
Question 10
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Whenever a stream of electrons collides with a stream of photons, in this collision, which of the following is not conserved?

A
Linear momentum

B
Total energy

C
No. of photons

D
No. of eletrons

Solution

When a stream of electrons collides with a stream of photons, then it may be possible that after the collisions, the respective velocity of some of the electrons will get changed, either it can increase or decrease depending on the orientation of the collision. But the number of electrons will remain the same. Moreover, the photons are absorbed by electrons to change their velocities.
Also total energy and linear momentum is always conserved according to the conservation laws.
Hence, number of photons is not conserved.

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Atoms Test - 11

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Atoms Test - 11

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Question 1

$$\displaystyle \frac{1}{\mathrm{n}^{2}}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

$$\displaystyle \frac{1}{\mathrm{n}}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

$$\displaystyle \mathrm{n}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

$$\displaystyle \mathrm{n}^{2}(\frac{\mathrm{h}^{2}}{8\pi^{2}\mathrm{I}})$$

Solution

Question 2

$$2.4\times 10^{-10}\mathrm{m}$$

$$1.9\times 10^{-10}\mathrm{m}$$

$$1.3\times 10^{-10}\mathrm{m}$$

$$4.4\times 10^{-11}\mathrm{m}$$

Solution

Question 3

$$2.85\ fm$$

$$3.03\ fm$$

$$3.42\ fm$$

$$3.80\ fm$$

Solution

Question 4

$$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}$$

$$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}$$

$$\mathrm{r}_{\mathrm{n}}\propto \mathrm{n}^{2}$$

$$\mathrm{r}_{\mathrm{n}}\propto 1/\mathrm{n}^{2}$$

Solution

Question 5

Singly ionised helium atom $$(He^+)$$

Deuteron atom

Singly ionised neon atom $$(Ne^+)$$

Hydrogen atom

Solution

Question 6

$$\dfrac { 1 }{ m } $$

$$\dfrac { 1 }{ \sqrt { m } } $$

$$\dfrac { 1 }{ { m }^{ 2 } } $$

$$m$$

Solution

Question 7

$$\gamma$$ rays

x rays

Infra - red rays

Ultraviolet rays

Solution

Question 8

$$10$$

$$25$$

$$50$$

$$75$$

Solution

Question 9

$$3\rightarrow 2$$

$$4\rightarrow 2$$

$$4\rightarrow 3$$

$$2\rightarrow 1$$

Solution

Question 10

Linear momentum

Total energy

No. of photons

No. of eletrons

Solution

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