Atoms Test

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Atoms Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

In hydrogen like atom electron makes transition from an energy level with quantum number $$n$$ to another with quantum number $$(n-1)$$. If $$n>> 1$$, the frequency of radiation emitted is proportional to :

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

B
$$\cfrac { 1 }{ { n }^{ 3/2 } } $$

C
$$\cfrac { 1 }{ { n }^{ 3 } } $$

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

Solution

$$\begin{array}{l} E=-13.6\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)=\frac{h c}{\lambda e} \\ -13.6\left(\frac{1}{n^{2}}-\frac{1}{(n-1)^{2}}\right)= \frac{h v}{e} \\ x-\frac{13 \cdot 6}{e}\left(\frac{n^{2}-n^{2}-1+2 n}{n^{2}(n-1)^{2}}\right)=h v \\ \frac{2 n-1}{n^{2}(n-1)^{2}}\times{v} \end{array} $$

$$V=\frac{2 n-1}{n^{2}\left(n^{2}+1-2 n\right)}$$
$$v_{i} \times \frac{2 n-1}{n^{4}+n^{2}-2 n^{3}} \ldots .$$
$$\therefore n>>1$$
$$v \times \frac{2 n}{n^{4}} \times \frac{2}{n^{3}} \times \frac{1}{n^{3}}$$
$$v \times \frac{1}{n^{3}}$$
Correct C
Question 2
1 / -0

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the $$n$$th orbital will be proportional to

A
$${ n }^{ 2 }$$

B
$$n$$

C
$${ n }^{ 1/2 }$$

D
$${ n }^{ 1/4 }$$

Solution

$$qvB=\cfrac { m{ v }^{ 2 } }{ r } ;mvr=\cfrac { nh }{ 2\pi } $$
$$\therefore \cfrac { qBr }{ m } =\cfrac { nh }{ 2\pi mr } $$
$$\therefore r\propto { n }^{ 1/2 }\quad $$
Question 3
1 / -0

The wavelengths involved in the spectrum of deuterium $$\left( _{ 1 }^{ 2 }{ D } \right) $$ are slightly different from that of hydrogen spectrum, because

A
size of the two nuclei are different

B
nuclear forces are different in the two cases

C
masses of the two nuclei are different

D
attraction between the electron and the nucleus is different in the two cases

Solution

The wavelengths involved in the spectrum of deuterium D are slightly different from the hydrogen spectrum because masses of the two nuclei are different.
Question 4
1 / -0

The gyromagnetic ratio of an electron in an $$H-$$atom, according to Bohr model, is:-

A
Independent of which orbit it is in

B
Negative

C
Positive

D
Increase with the quantum number $$n$$

Solution

$$\begin{array}{l}\text { we know that the - } \\\text { Gyromagnetic Ratio = } \frac{q}{2 m} \\\text { Since, the electron carries a negative charge. } \\\text { Hence, the gyromagnetic ratio will be negative .}\end{array}$$
Question 5
1 / -0

The ionization energy of hydrogen atom is $$13.6\ eV$$. Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy $$12.1\ eV$$. How many spectral lines will be emitted by the hydrogen atoms?

A
one

B
two

C
three

D
four

Solution

Energy of electron in hydrogen atom in ground state=-13.5eV.

When a photon of energy of 12.1eV is radiated on atom, the energy gained by the electron is given by :

$$E=(-13.6)+12.1=(-1.5)eV$$

And we know that energy of an excited electron is given by,

$$E=\dfrac { -13.6 }{ { n }^{ 2 } } (n:number\quad of\quad orbit)$$

$$\Rightarrow -1.5=\dfrac { -13.6 }{ { n }^{ 2 } } \\ \Rightarrow n=3$$

Therefore, when a photon of 12.6eV is radiated electron jump to 3 orbit.
Question 6
1 / -0

Directions For Questions

A uniform magnetic field $$B$$ exists in a region. An electron is given velocity perpendicular to the magnetic field. Assuming Bohrs quantization rule for angular momentum

...view full instructions

Calculate the radius of the nth orbit

A
$$\sqrt{\dfrac{nh}{2\pi e B}}$$

B
$$\sqrt{\dfrac{nheB}{2\pi}}$$

C
$$\sqrt{\dfrac{nhe}{2\pi B}}$$

D
$$\sqrt{\dfrac{nhB}{2\pi e}}$$

Solution

According to Bohr's quantization rule
$$mvr=\dfrac{nh}{2\pi}$$
'r' is less when 'n' has least value i.e. 1
or $$mv=\dfrac{nh}{2\pi r}.....(1)$$
Again, $$r=\dfrac{mv}{qB}$$
or $$mv=rqB.....(2)$$
From (1) and (2)
$$rqB=\dfrac{nh}{2\pi r}$$ [q=e]
$$\Rightarrow r^2=\dfrac{nh}{2\pi eB}\Rightarrow r=\sqrt{\dfrac{h}{2\pi eB}}$$ [Here, n=1]
For the radius of nth orbit, $$r=\sqrt{\dfrac{nh}{2\pi eB}}$$
Question 7
1 / -0

Consider Bohr's theory for hydrogen atom. The magnitude of angular momentum, orbit radius and frequency of the electron in $$n^th$$ energy state in a hydrogen atom are, $$r$$ and $$f$$ respectively. find out the value of $$X$$, if ($$frl$$) is directly proportional to $$n^x$$.

A
1

B
0

C
2

D
3

Solution

We know that-

$$r\propto n^2$$

$$v\propto \dfrac{1}{n}$$

and $$l=\dfrac{nh}{2\pi}\implies l\propto n$$

$$f=\dfrac{1}{T}=\dfrac{v}{2\pi r}$$

$$\implies f\propto \dfrac{1}{n^3}$$

Hence, $$frl \propto n^0$$

Hence, $$x=0$$
Question 8
1 / -0

Electron in a hydrogen atom is replaced by an identically charged particle muon with mass $$207$$ times that of electron . Now the radius of K shell will be

A
$$2.56 \times 10^{-3}\mathring {A}$$

B
$$109.7 \mathring{A}$$

C
$$1.21 \times 10^{-3}\mathring{A}$$

D
$$22174.4 \mathring {A}$$

Solution
Question 9
1 / -0

There are some atoms in ground energy level and rest of all in some upper energy level in a gas of identical hydrogen like atoms. The atoms jump to higher energy level by absorbing monochromatic light of photon energy $$1.89\ eV$$ and this emit radiations of $$3$$ photons energies. Find the principal quantum number of initially excited level.

A
$$3$$

B
$$4$$

C
$$2$$

D
$$5$$

Solution
Question 10
1 / -0

Ionization potential of hydrogen is $$13.6$$ volt. If it is excited by a photon of energy $$12.1eV$$, then the number of lines in the emission spectrum will be:

A
$$2$$

B
$$3$$

C
$$4$$

D
$$5$$

Solution

In this case, the energy of H-atom after it absorbs photon of energy 12.1eV will be $$=-13.6eV+12.1eV=-1.5eV$$
Since,
$$E_n=-R_H\left(\frac{1}{n^2}\right)$$
$$n^2=\frac{-R_H}{E_n}=\frac{-2.18\times 10^{-18}}{-1.5\times 1.6\times 10^{-19}}=9.08$$
$$n=3$$

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Atoms Test - 71

Result

Atoms Test - 71

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Question 1

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

$$\cfrac { 1 }{ { n }^{ 3/2 } } $$

$$\cfrac { 1 }{ { n }^{ 3 } } $$

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

Solution

Question 2

$${ n }^{ 2 }$$

$$n$$

$${ n }^{ 1/2 }$$

$${ n }^{ 1/4 }$$

Solution

Question 3

size of the two nuclei are different

nuclear forces are different in the two cases

masses of the two nuclei are different

attraction between the electron and the nucleus is different in the two cases

Solution

Question 4

Independent of which orbit it is in

Negative

Positive

Increase with the quantum number $$n$$

Solution

Question 5

one

two

three

four

Solution

Question 6

$$\sqrt{\dfrac{nh}{2\pi e B}}$$

$$\sqrt{\dfrac{nheB}{2\pi}}$$

$$\sqrt{\dfrac{nhe}{2\pi B}}$$

$$\sqrt{\dfrac{nhB}{2\pi e}}$$

Solution

Question 7

1

0

2

3

Solution

Question 8

$$2.56 \times 10^{-3}\mathring {A}$$

$$109.7 \mathring{A}$$

$$1.21 \times 10^{-3}\mathring{A}$$

$$22174.4 \mathring {A}$$

Solution

Question 9

$$3$$

$$4$$

$$2$$

$$5$$

Solution

Question 10

$$2$$

$$3$$

$$4$$

$$5$$

Solution

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