Atoms Test

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Atoms Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The photon energy in units of eV for electromagnetic waves of wavelength $$2 cm$$ is:

A
$$2.5 10^{-19}$$

B
$$5.2 10^{-16}$$

C
$$3.2 10^{-16}$$

D
$$6.2 10^{-5}$$

Solution

Given,
$$\lambda=2cm=0.02m$$
$$h=6.62\times 10^{-34} kgm^2/s$$
$$c=3\times 10^8m/s$$
Photon energy, $$E=\dfrac{hc}{\lambda}$$
$$E=\dfrac{6.62\times 10^{-34}\times 3\times 10^8}{0.02}$$
$$E=993\times 10^{-26}J$$
Photon energy in unit of $$eV$$,
$$E=\dfrac{993\times 10^{-26}}{1.6\times 10^{-19}}eV$$
$$E=6.20\times 10^{-5}eV$$
Question 2
1 / -0

An electron with energy $$12.09$$ eV strikes hydrogen atom in ground state and gives its all energy to the hydrogen atom. Therefore hydrogen atom is excited to __________ state.

A
Fourth

B
Third

C
Second

D
First

Solution

We know that energy of Hydrogen atom in it's orbit is given by formula $$\dfrac{-13.6}{n^2} eV$$.

Initial Energy of Electron = $$-13.6 eV$$ (Given that atom in i's ground state i.e n=1)

Energy after striking = $$(-13.6 + 12.09) eV = -1.51 eV$$

$$\Rightarrow$$ $$\dfrac{-13.6}{n^2} eV = 1.51eV$$ $$\Rightarrow$$ $$n^2= 9$$ $$\Rightarrow$$ $$ n=3.$$

So, hydrogen atom is excited to third state.

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

The radius of second orbit in an atom of hydrogen is R. What is the radius in third orbit.

A
$$3R$$

B
$$2.25R$$

C
$$9R$$

D
$$\dfrac{R}{3}$$

Solution

$$r_n=0.529\dfrac{n^{2}}{Z} \overset{o}{A}$$ where $$r_n$$ is radius of $$n^{th}$$ orbit.
$$n=$$ number of the orbit.
$$Z=$$ atomic number.
Given:
$$Z=1$$ (Hydrogen orbit)
$$R=0.529\times \dfrac{(2)^{2}}{1}=4\times 0.529$$
Now let radius of $$3^{th}$$ orbit be x.
$$x=0.529\dfrac{n^{2}}{Z} \overset{o}{A}=0.529\dfrac{(3)^{2}}{Z}\overset{o}{A}=0.529\times \dfrac{9}{1} \overset{o}{A}=0.529\times\dfrac{4}{4} \times \dfrac{9}{1} \overset{o}{A}=R\times \dfrac{9}{4}=2.25R$$
Hence the correct option is (B).
Question 4
1 / -0

In the geiger-marsden experiment what percentage of the $$\alpha$$ particles were deflected by an angle more than $$1^o$$.

A
$$< 0.15 %$$

B
$$< 15 %$$

C
$$> 99.85 %$$

D
$$100 %$$

Solution

The Geiger-Marsden experiment, also called as the Rutherford's gold foil experiment were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where all of its positive charge and most of its mass are concentrated. They deduced this by measuring how an alpha particle beam is scattered when it strikes a thin metal foil. In this experiment they observed that only less than $$15$$% of th alpha partils
Question 5
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What is alpha-particle?

A
The nucleus of hydrogen atom

B
The nucleus of deuterium

C
A fast moving proton

D
The nucleus of helium atom

Solution

$$\begin{array}{l}\text { Alpha particle is the nucleus of a } \\\text { helium atom } ^4_2 \mathrm{He} \text { . }\end{array}$$
Question 6
1 / -0

In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by

A
increasing the potential difference between the anode and filament

B
increasing the filament current

C
decreasing the filament current

D
decreasing the potential difference between the anode and filament

Solution

when the potential increases, the Potential difference between the electrodes increase and the flow of electrons also increases.
Hence A is correct.
Question 7
1 / -0

If the series limit of the Balmer series for hydrogen is 2700 Angstrom. Calculate the atomic no. of the element which gives X-ray wavelength of $$K_\alpha$$ line as 1.0 Angstrom.

A
z = 21

B
z = 31

C
z = 61

D
z = 5

Solution

wavelength of Balmer sines $$=3646 \overset { o }{ A } $$
Rydbergs formula :
$$v-=dfrac{1}{\lambda}=R\left(\dfrac{1}{2^{2}}-\dfrac{1}{\infty^{2}}\right)$$
$$\bar{v}=\dfrac{R}{4}$$
$$\therefore R=\dfrac{4}{\lambda}=\dfrac{4}{3646}\times 10^{10}m^{-1}$$
wavelength of $$K_{a}$$sins :
$$\bar{v}=\dfrac{1}{\lambda}=R(2-1)^{2}\left(\dfrac{1}{1^{2}}-\dfrac{1}{h^{2}}\right)$$
$$n=\infty$$
$$\therefore \bar{v}=\dfrac{1}{\lambda}=R (2-1)^{2}$$
$$\therefore 2-1=\sqrt{911.5}$$
$$=30.2$$
$$z=31$$
Question 8
1 / -0

Cathode rays were discovered by

A
Maxwell Clerk James

B
Heinrich Hertz

C
William Crookes

D
J. J. Thomson

Solution

J.J Thomson
Question 9
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The ionization energy of a hydrogen like Bohr atom is $$4$$ Rydbergs. (i) What is the wavelength of the radiation emitted when an electron jumps from the first excited state to the ground state? (ii) What is the radius of the first orbit for this atom?

A
$$300\overset {\circ}{A}, 2.65\times 10^{-11}m$$.

B
$$600\overset {\circ}{A}, 2.65\times 10^{-11}m$$.

C
$$300\overset {\circ}{A}, 4\times 10^{-11}m$$.

D
$$600\overset {\circ}{A}, 4\times 10^{-11}m$$.

Solution
Question 10
1 / -0

A double ionized lithium atom is hydrogen-like with atomic number $$3$$.
(a) Find the wavelength of the radiation required to excite the electron in $$Li++$$ from the first to the third Bohr orbit. (Ionization energy of the hydrogen atom equals $$13.6\ eV$$)\
(b) How many spectral lines are observed in the emission spectrum of the above excited system?

A
$$114.7\ A^{\circ}$$, Three lines.

B
$$113.7\ A^{\circ}$$, four lines.

C
$$115.7\ A^{\circ}$$, two lines.

D
$$116.7\ A^{\circ}$$, one lines.

Solution

formula used $$=-13.6\dfrac{z^{2}}{n^{2}}eV$$
now, for $$n=1, E_{i}=-122.4\ eV$$
for $$n=3, E_{j}=-13.6\ eV$$
now, photon energy required $$E_{p}=E_{j}-E_{i}$$
$$\dfrac{h o}{\lambda}=\left\{-13.6-(-122.4)\right\}eV$$
$$\lambda=\dfrac{12400\ eV}{108.8\ eV}-A$$
$$\lambda=113.97\ A$$
spectral lines are $$3\rightarrow 2\rightarrow 1\rightarrow 0$$
$$\therefore$$ four lines
$$\therefore$$ Option $$(B)113.7\ A^{o}$$, four lines is correct option

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Atoms Test - 47

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Atoms Test - 47

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Question 1

$$2.5 10^{-19}$$

$$5.2 10^{-16}$$

$$3.2 10^{-16}$$

$$6.2 10^{-5}$$

Solution

Question 2

Fourth

Third

Second

First

Solution

Question 3

$$3R$$

$$2.25R$$

$$9R$$

$$\dfrac{R}{3}$$

Solution

Question 4

$$< 0.15 %$$

$$< 15 %$$

$$> 99.85 %$$

$$100 %$$

Solution

Question 5

The nucleus of hydrogen atom

The nucleus of deuterium

A fast moving proton

The nucleus of helium atom

Solution

Question 6

increasing the potential difference between the anode and filament

increasing the filament current

decreasing the filament current

decreasing the potential difference between the anode and filament

Solution

Question 7

z = 21

z = 31

z = 61

z = 5

Solution

Question 8

Maxwell Clerk James

Heinrich Hertz

William Crookes

J. J. Thomson

Solution

Question 9

$$300\overset {\circ}{A}, 2.65\times 10^{-11}m$$.

$$600\overset {\circ}{A}, 2.65\times 10^{-11}m$$.

$$300\overset {\circ}{A}, 4\times 10^{-11}m$$.

$$600\overset {\circ}{A}, 4\times 10^{-11}m$$.

Solution

Question 10

$$114.7\ A^{\circ}$$, Three lines.

$$113.7\ A^{\circ}$$, four lines.

$$115.7\ A^{\circ}$$, two lines.

$$116.7\ A^{\circ}$$, one lines.

Solution

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