Atoms Test

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Atoms Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

Let $$v_{1}$$ be the frequency of the series limited of the Lyman series, $$v_{2}$$ be the frequency of the first line of the Lyman series, and $$v_{3}$$ be the frequency of the series limited of the Balmer series. Then:

A
$$v_{1}-v_{2}=v_{3}$$

B
$$v_{2}-v_{1}=v_{3}$$

C
$$v_{3}=\frac{1}{2} (v_{1}+v_{3})$$

D
$$v_{1}+v_{2}=v_{3}$$

Solution

From Rydberg's relation:
$$\dfrac{1}{\lambda}=R_H(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$$
$$\implies \nu=R_Hc(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$$
First line of lyman series is from transition $$2\rightarrow 1$$
Series limit of lyman is from transition $$\infty\rightarrow 1$$
Series limit of balmer is from transition $$\infty\rightarrow 2$$
$$\nu_1=R_Hc(1-\dfrac{1}{\infty^2})$$
$$\nu_2=R_Hc(1-\dfrac{1}{2^2})$$
$$\nu_3=R_Hc(\dfrac{1}{2^2}-\dfrac{1}{\infty^2})$$
$$\nu_1-\nu_2=R_Hc(\dfrac{1}{2^2}-\dfrac{1}{\infty^2})=\nu_3$$
Question 2
1 / -0

If $$ A_{n}$$ is the area enclosed in the $$ n$$th orbit in a hydrogen atom then the graph log $$ \left ( \frac{A_{n}}{A_{1}} \right )$$ against log $$ n$$:

A
will have slope 2 (a straight line)

B
will have slope 4 (a straight line)

C
will be a monotonically increasing non-linear curve

D
will be a circle

Solution

Concept attached
$$r_n=Cn^2$$
$$ r_1 = C $$
$$r_n= r_1 n^2$$
$$A_n= \pi r_n ^2 = \pi r_1 ^2 n^4$$
$$ log(\cfrac{A_n}{A_1}) = 4 logn +K$$
Slope =4, Linear graph.
Question 3
1 / -0

The quantum number n of the state finally populated in $$He^{+}$$ ions is

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

$$\Delta E_{h}=\dfrac{3}{4}\times 13.6\ eV =$$ Energy released by H atom.

Let He$$^{+}$$go to nth state.

So energy required, $$\Delta E_{he}= 13.6 \times 4\left ( \dfrac{1}{4}-\dfrac{1}{n^{2}} \right )$$eV
$$\Delta E_{he}=\Delta E_{h}$$

$$\Rightarrow \dfrac{3}{4}\times 13.6=13.6\times 4\left ( \dfrac{1}{4}-\dfrac{1}{n^{2}} \right )$$

$$\Rightarrow$$ $$n=4$$
Question 4
1 / -0

In the figure six lines of emission spectrum are shown. Which of them will be absent in the absorbtion spectrum.

A
1,2,3

B
1,4,6

C
4,5,6

D
1,2,3,4,5,6

Solution

Here all electrons are coming from the higher state to lower state. So in the absorption spectrum there should be lower to higher state.
So, $$1,2,3,4,5,6$$ are the absent in the absorption spectrum.
Question 5
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Check the correctness of the following statements about the Bohr Model of hydrogen atom:
$$(i)$$ The acceleration of the electron in $$n=2$$ orbit is more than that in $$n=1$$ orbit.
$$(ii)$$ The angular momentum of the electron in $$n=2$$ orbit is more than that in $$n=1$$ orbit.
$$(iii)$$ The KE of the electron in $$n=2$$ orbit is less than that in $$n=1$$ orbit

A
All the statements are correct

B
Only $$(i)$$ and $$(ii)$$ are correct

C
Only $$(ii)$$ and $$(iii)$$ are correct

D
Only $$(iii)$$ and $$(i)$$ are correct

Solution

Angular momentum, $$L=mvr_n=\dfrac{nh}{2\pi}\propto n$$
Hence, (ii) is correct.
Velocity of $$2^{nd}$$ orbit is less so kinetic energy is also less.
Hence, (iii) is correct.
The radius of $$2^{nd}$$ orbit is $$4$$ times the radius of $$1^{st}$$ orbit.
Thus, acceleration $$(v^2/r)$$ in $$n=2$$ is less than the $$n=1$$.
Hence, (i) is incorrect.
Question 6
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Directions For Questions

The physics teacher in class explained Bohr model of atom. He told that Bohr assumed that angular momentum is quantized. A student very politely wanted to confirm whether the same phenomenon can be applied elsewhere, for example, in planetary system or other systems in which some particles revolve about the other. Another student was anxious to know can de-Broglie relation be applied. Physics teacher explained so beautifully about classical and quantum mechanical nature.

...view full instructions

The absorption spectrum of $$H$$-atom is found in _______.

A
Lyman series

B
Balmer series

C
Paschen series

D
In all cases

Solution

The Lyman series is a Hydrogen spectral series of transitions and resulting ultraviolet emission lines of the hydrogen atom as an electron goes from $$n = 2$$ to $$n = 1$$ (where n is the principal quantum number) the lowest energy level of the electron.
Question 7
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Directions For Questions

A $$20 kg$$ satellite circles the earth once every $$2h$$ in an orbit having a radius of $$8060 km$$. Assume that Bohrs postulates are valid even in the satellite motion as for electrons in Hydrogen atom. Combine Newton's law of gravitation and Bohr's angular momentum to find radius of the orbit as a function of principle Quantum number.

...view full instructions

Find the value of principal quantum number n corresponding 20 kg satellite.

A
$$1.6\times 10^{45}$$

B
$$1.6\times 10^{8}$$

C
$$7.6\times 10^{5}$$

D
$$8.3\times 10^{11}$$

Solution

We know that:
$$nh=mvR$$, where $$R= $$ $$8060\times 10^3 m$$
Also, $$v=\sqrt{\dfrac{GM}{R}}$$, $$M =$$ mass of earth $$=6\times 10^24kg$$

$$n = \displaystyle \frac{m}{h} \sqrt{\displaystyle \frac{GM}{R}} R = \displaystyle \frac{20}{6.62\times 10^{-34}}\sqrt{6.67\times 10^{-11}\times 6\times 10^{24}\times 6400\times 10^3}\ \ 8060\times 10^3 $$

$$= \displaystyle \frac{20\times 10^{34}}{6.62}\times 10^9\times 8\times \sqrt {40} \\ = 1.6\times 10^{45}$$
Question 8
1 / -0

The figure above represents some of the lower energy levels of the hydrogen atom in simplified form.
If the transition of an electron from $$E_4$$ to $$E_2$$ were associated with the emission of blue light, which one of the following transitions could be associated with the emission of red light?

A
$$E_4$$ to $$E_1$$

B
$$E_3$$ to $$E_1$$

C
$$E_3$$ to $$E_2$$

D
$$E_1$$ to $$E_3$$

Solution

Wavelength of blue is less than wavelength of red.
For transition into $$n=2$$ (Balmer series), wavelength is inversely proportional to n
Hence, for wavelength to increase, $$n$$ should be less than $$4$$
Therefore, $$n=3$$
Question 9
1 / -0

Directions For Questions

A $$20 kg$$ satellite circles the earth once every $$2h$$ in an orbit having a radius of $$8060 km$$. Assume that Bohrs postulates are valid even in the satellite motion as for electrons in Hydrogen atom. Combine Newton's law of gravitation and Bohr's angular momentum to find radius of the orbit as a function of principle Quantum number.

...view full instructions

Find the radius of first two allowed orbits.

A
$$2.5 \times 10^{-84}m, 10^{-83}m$$

B
$$400 km, 1600 km$$

C
$$200 m, 800 m$$

D
$$3 km, 12 km$$

E
none

Solution

$$ mvR=nh$$
$$V_o = \sqrt{\cfrac{GM}{R}}$$
Substitute:
$$m\sqrt{\dfrac{GM}{R}}R=nh\\ R = \cfrac{n^2h^2}{m^2GM}$$
$$R = n^2 k$$

$$ k = 1.6 \times 10^{-84} $$ m
$$ R_1 $$ = 1.6 $$ \times 10^{-84} m $$
$$ R_2 = 6.4 \times 10^{-84} $$ m
Question 10
1 / -0

In a hydrogen atom, electron is in the $$n^{th}$$ excited state. It comes down to the first excited state by emitting $$10$$ different wavelengths. The value of $$n^{th}$$ is

A
$$6$$

B
$$7$$

C
$$8$$

D
$$9$$

Solution

Total no. of levels $$= n-1$$
$$ n \rightarrow 2$$
Put $$n' = n-1 $$ in the formula attached
$$ \cfrac{(n-1) (n-2)}{2} = 10$$
$$ (n-1)(n-2) = 20$$
$$\Rightarrow n=6$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 67

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Atoms Test - 67

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SHARING IS CARING

Question 1

$$v_{1}-v_{2}=v_{3}$$

$$v_{2}-v_{1}=v_{3}$$

$$v_{3}=\frac{1}{2} (v_{1}+v_{3})$$

$$v_{1}+v_{2}=v_{3}$$

Solution

Question 2

will have slope 2 (a straight line)

will have slope 4 (a straight line)

will be a monotonically increasing non-linear curve

will be a circle

Solution

Question 3

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 4

1,2,3

1,4,6

4,5,6

1,2,3,4,5,6

Solution

Question 5

All the statements are correct

Only $$(i)$$ and $$(ii)$$ are correct

Only $$(ii)$$ and $$(iii)$$ are correct

Only $$(iii)$$ and $$(i)$$ are correct

Solution

Question 6

Lyman series

Balmer series

Paschen series

In all cases

Solution

Question 7

$$1.6\times 10^{45}$$

$$1.6\times 10^{8}$$

$$7.6\times 10^{5}$$

$$8.3\times 10^{11}$$

Solution

Question 8

$$E_4$$ to $$E_1$$

$$E_3$$ to $$E_1$$

$$E_3$$ to $$E_2$$

$$E_1$$ to $$E_3$$

Solution

Question 9

$$2.5 \times 10^{-84}m, 10^{-83}m$$

$$400 km, 1600 km$$

$$200 m, 800 m$$

$$3 km, 12 km$$

none

Solution

Question 10

$$6$$

$$7$$

$$8$$

$$9$$

Solution

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