Atoms Test

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Atoms Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

The quantum number corresponding to orbit of diameter 0.0001 mm in hydrogen atom will be nearly (Given that the radius of orbit with n = 1 is $$0.51\times 10^{-10}$$):

A
39

B
31

C
9

D
49

Solution

Radius of orbit $$=\dfrac{0.0001}{2}=5\times 10^{-5}mm=5\times 10^{-8}m$$
Radius of $$n^{th} $$ orbit is given by :- [from Bohr's model]
$$r_n=r_0n^2$$
where $$r_n=$$ radius of $$1st$$ orbit
$$=0.51\times 10^{-10}m$$
n: Quantum number of orbit
$$\Rightarrow n^2=\dfrac{r_n}{r_0}=\dfrac{5\times 20^{-8}}{0.51\times 10^{-1}}$$
$$\Rightarrow n^2\simeq 2381 $$
So,$$n=\sqrt{2381 }\simeq49$$
Question 2
1 / -0

How many times does the electron go round the first Bohr orbit in a second?

A
$$6.57 \times 10 ^ { 5 }$$

B
$$6.57 \times 10 ^ { 10 }$$

C
$$6.57 \times 10 ^ { 13 }$$

D
$$6.57 \times 10 ^ { 15 }$$

Solution

Frequency $$(\nu) = \dfrac{v}{2\pi r}$$

where, v is velocity of electron and r is radius of first Bohr orbit.

$$ \Rightarrow \nu = \dfrac{2.2 \times 10^6}{2 \pi (0.5287 \times 10^{-10} m) } = 6.57 \times 10^{15}$$

Therefore, D is correct option.
Question 3
1 / -0

The radius of the first Bohr orbit of a hydrogen atom is $$ 0.53\times 10^{-10}m$$.When an electron collides with this atom which is in its normal state, the radius of the electron orbit in the atom change to $$2.12 \times 10^{10}m$$.The value of the principal quantum number $$n$$ of the state to which it is excited is:

A
$$4$$

B
$$3$$

C
$$2$$

D
$$1$$

Solution

$$r_n=a_0n^2$$
$$2.12\times 10^{-19}=90n^2$$
$$=0.53n^2$$
$$n^2=4$$
$$n=2$$
This value of principle quantum number of excited state in $$2$$
Question 4
1 / -0

If energy required to remove one of the two electron from $$\text { He }$$ atom is $$29.5 eV,$$ then what is the value of energy required to convert a helium atom into $$\alpha -$$particle?

A
$$54.4eV$$

B
$$83.9eV$$

C
$$29.5eV$$

D
$$24.9eV$$

Solution

From a He-atom, containing $$ 2e^{-} $$ m ground state, one electron is
removed by proving 29.5 ev of energy, and remain $$ He^{+} $$
$$ He \overset{29.sev}{\rightarrow} He^{+}+e^{-} $$
$$ He^{+} $$ (single $$ e^{-} $$ species ) because uke Bohr H-atom having ground
state energy,
$$ E_{1} = -13.6\times \dfrac{z^{2}}{n^{2}}$$ , for $$ z = 2 (He)2n = 1 $$
$$ = -13.6\times \dfrac{z^{2}}{12} = -54.4ev $$
So, an addition 54.4ev should be given to convent $$ He^{+} $$ to
$$ \alpha -$$ partial (which is basically He nuclei, or $$ He^{++}$$).
$$ He^{+}\overset{54.4 ev}{\rightarrow} He^{++}+e^{-} $$
Net energy required $$ = 29.5+54.4 = 83.9ev $$
Question 5
1 / -0

Photoelectric effect was successfully explained by

A
Planck

B
Hallwash

C
Hertz

D
Einstein

Solution

Einstein successfully explained the photoelectric effect using the assumption $$E=h\nu $$. This assumption is given by Planck. Einstein showed that not only light is quantized but also atomic vibrations are quantized. He was able to explain this result by assuming that the oscillations of atoms are quantized according to $$E=nh\nu $$.
Question 6
1 / -0

The ratio of the speed of the electron in the first Bohr orbit of hydrogen and the speed of light is equal to then (where e, h and c have their usual meanings)

A
$$\dfrac{e^{12}}{2C\epsilon^{h}}$$

B
$$\dfrac{e^{2}}{2C\epsilon^{h}}$$

C
$$\dfrac{e^{3}}{2C\epsilon^{h}}$$

D
$$\dfrac{e^{4}}{2C\epsilon^{h}}$$

Solution

reed of an electron in $$n^{th}$$ orbit (for Hydrogen- like system) is given by :
$$v= \dfrac{KZl^{2}}{nh} [where\ K: \dfrac{1}{4 \pi \epsilon _{0}} \times h= \dfrac{h}{2\pi} ]$$
For first orbit of Hydrogen, $$n=1\ k\ z=1$$
$$\Rightarrow v_{1}= \dfrac{2 \pi e^{2}}{4 \pi \epsilon_{0} h}=\dfrac{e^{2}}{2 \epsilon_{0} h}.$$
Now, velocity of speed, $$=c$$, so Ratio :
$$\Rightarrow \dfrac{v_{1}}{c}=\dfrac{e^{2}}{2c \epsilon_{0} h }.$$
Question 7
1 / -0

The ratio of the velocity of an electron in the first Bohr's orbit of the hydrogen atom and the velocity of light is:

A
$$1:100$$

B
$$1:137$$

C
$$1:1000$$

D
$$1:10$$

Solution

In hydrogen atom the speed of electron is $$v=\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}nh}........(1)$$

Where, e is the charge of electron

n = 1 as first orbit

Dividing equation (1) by c

$$ \dfrac{v}{c}=\dfrac{{{e}^{2}}}{2{{\varepsilon }_{0}}ch} $$

$$ \dfrac{v}{c}=\dfrac{{{(1.6\times {{10}^{-19}})}^{2}}}{2\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}\times 6.6\times {{10}^{-34}}} $$

$$ \dfrac{v}{c}=\dfrac{1}{137} $$
Ratio is 1 : 137
Question 8
1 / -0

If an electron in a hydrogen atom has moved from $$n = 1$$ to $$n = 10$$ orbit, the potential energy of the system has:

A
increased

B
decreased

C
remained unchanged

D
become zero

Solution

Solution
In Bohrs atomic at model of H -atom
$$ U = (-13.6ev) \dfrac{z^{2}}{n^{2}} = \dfrac{(-13.6ev(1)^{2})}{n^{2}}$$
$$ \Rightarrow U = \dfrac{-13.6ev}{n^{2}}$$
For $$ n = 1,$$
$$ U = -13.6ev$$
For n = 10,
$$U = -0.136 ev$$
As it has got less negative, it
has been increased.
Question 9
1 / -0

The ratio of the orbit of the $$1st $$ three radii in an atom of hydrogen is

A
$$1:2:3$$

B
$$3:2:1$$

C
$$1:4:9$$

D
$$9:4:1$$

Solution

$$\begin{array}{l}\text { Radii of nth orbit in an hydrogen } \\\text { atom is proportional to n }^{2} \\\therefore \text { Required ratio is } \\\qquad \begin{array}{l}(1)^{2}:(2)^{2}:(3)^{2} \\=1: 4: 9\end{array}\end{array}$$
Question 10
1 / -0

Ratio of series limit of Brackett and Pfund series of hydrogen ato spectrum is

A
$$\dfrac{1}{4}$$

B
$$\dfrac{4}{9}$$

C
$$\dfrac{9}{16}$$

D
$$\dfrac{16}{25}$$

Solution

$$\begin{array}{c}{\text { Bracket series }} \text { n }_{1}=3, n_{2}=4,5,\ldots \\\dfrac{1}{\text { series limit }}\propto \left[\dfrac{1}{3^{2}}-\dfrac{1}{n_2^2}\right] \\\propto \dfrac{1}{9}\end{array}$$

$$\begin{array}{l}\text { Pfund series } \quad n_{1}=4, n_{2}=5,6,\ldots \\\dfrac{1}{\text { series limit }}\propto \left[\dfrac{1}{4^{2}}-\dfrac{1}{(n_2)^{2}}\right] \propto \dfrac{1}{16} \\\text { Ratio of series limit }=9/16\end{array}$$

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Re-Attempt Weekly Quiz Competition

Atoms Test - 53

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Atoms Test - 53

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SHARING IS CARING

Question 1

39

31

9

49

Solution

Question 2

$$6.57 \times 10 ^ { 5 }$$

$$6.57 \times 10 ^ { 10 }$$

$$6.57 \times 10 ^ { 13 }$$

$$6.57 \times 10 ^ { 15 }$$

Solution

Question 3

$$4$$

$$3$$

$$2$$

$$1$$

Solution

Question 4

$$54.4eV$$

$$83.9eV$$

$$29.5eV$$

$$24.9eV$$

Solution

Question 5

Planck

Hallwash

Hertz

Einstein

Solution

Question 6

$$\dfrac{e^{12}}{2C\epsilon^{h}}$$

$$\dfrac{e^{2}}{2C\epsilon^{h}}$$

$$\dfrac{e^{3}}{2C\epsilon^{h}}$$

$$\dfrac{e^{4}}{2C\epsilon^{h}}$$

Solution

Question 7

$$1:100$$

$$1:137$$

$$1:1000$$

$$1:10$$

Solution

Question 8

increased

decreased

remained unchanged

become zero

Solution

Question 9

$$1:2:3$$

$$3:2:1$$

$$1:4:9$$

$$9:4:1$$

Solution

Question 10

$$\dfrac{1}{4}$$

$$\dfrac{4}{9}$$

$$\dfrac{9}{16}$$

$$\dfrac{16}{25}$$

Solution

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