Structure of Atom Test

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Structure of Atom Test - 71

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Question 1
1 / -0

Bohr's theory can also be applied to the ions like :

A
$$He^+$$

B
$$Li^{2+}$$

C
$$Be^{3+}$$

D
All of the these

Solution

Bohr theory is applicable to one-electron systems. Thus it is applicable to hydrogen or hydrogen-like atoms. Hydrogen-like atoms mean atoms/ions having only one electron. Such as $$He^+, Li^{2+},Be^{3+}$$ etc.
Question 2
1 / -0

An element with a mass number of $$81$$ contains $$31.7\%$$ more neutrons as compared to protons. Identify the element.

A
Sc

B
Ba

C
Br

D
I

Solution

An element with a mass number of $$81$$ contains $$31.7\%$$ more neutrons as compared to protons. The element is bromine.

Let P be the atomic number.
Number of protons $$ \displaystyle =P$$

Number of neutrons $$ \displaystyle =\dfrac { 100+ 31.7 }{ 100 } \times P=1.317P$$

The mass number is 81. It is the sum of the number of protons and the number of neutrons.

$$ \displaystyle 81 = P+1.317P=2.317P$$

$$ \displaystyle \Rightarrow P=35$$.

It is atomic number of bromine.
Question 3
1 / -0

How many maximum spectral lines are possible if the electron is present in the 4th shell and only two atoms are present in the sample?

A
6

B
4

C
3

D
2

Solution

The maximum number spectral lines possible is 4. 3 spectral lines are obtained for the transition $$4\to 3\to 2\to 1$$ and one spectral line is obtained for any one of the following transition $$4\to 1$$ or $$4\to 3\to 1$$ or $$4\to 2\to 1.$$
Question 4
1 / -0

If the radial probability curve indicates $$2s$$ orbital, the distance between the peak points X and Y is?

A
$$2.07\overset{o}{A}$$

B
$$1.59\overset{o}{A}$$

C
$$0.53\overset{o}{A}$$

D
$$2.12\overset{o}{A}$$
Question 5
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 6
1 / -0

Time taken for an electron to complete one in revolution in Bohr orbit of hydrogen atom is :

A
$$\frac{4\pi^2 mr^2}{nh}$$

B
$$\frac{nh}{4\pi^2 mr}$$

C
$$\frac{2\pi mr}{n^2h^2}$$

D
$$\frac{h}{2\pi mr}$$

Solution

Hint: We know that distance is a multiple of velocity and time.
Correct Answer: Option (A).

Explanation:

The time taken by an electron to complete one revolution of Bohr's orbit of the hydrogen atom $$=\dfrac{distance}{velocity}$$

Bohr's orbit is circular. So, the distance $$=2\pi r$$. .....$$(i)$$

Where $$r$$ is the radius of the Bohr's orbit.

Again we know that $$mvr=\dfrac{nh}{2\pi}$$

$$v=\dfrac{nh}{2\pi mr}$$ ....$$(ii)$$
$$t=\dfrac {distance}{velocity}$$
From the equations $$(i)$$ and $$(ii)$$, we got

$$t=\dfrac{2\pi r}{\dfrac{nh}{2\pi mr}}$$

$$\therefore t=\dfrac{4{\pi}^2mr^2}{nh}$$.

Final Answer: Time taken by an electron to complete one revolution in Bohr's orbit of the hydrogen atom is $$\dfrac{4{\pi}^2mr^2}{nh}$$.
Question 7
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,
$$E_n=-R_H\left(\frac{1}{n^2}\right)$$
$$n^2=\frac{-R_H}{E_n}=\frac{-2.18\times 10^{-18}}{-1.5\times 1.6\times 10^{-19}}=9.08$$
$$n=3$$
Question 8
1 / -0

The figure shown the wavelength spectrum of x-rays produced when $$50 \ KeV$$ electrons strike a molybdenum target. The $${ K }_{ \alpha }$$ line on the figure is represent by a number:

A
$$1$$

B
$$2$$

C
$$3$$

D
none

Solution

$$K_{\alpha}$$ refers to the peak due to $$e^{-}$$ transition from $$n=1$$ orbit
Ans is $$(A)$$
Question 9
1 / -0

A substance can be identified because its chemical environment often shows up like a finger print in a spectrum. This fine region of finger prints of the chemical environment of a chemical compound is created by:

A
infrared radiation

B
ultraviolet light

C
nuclear magnetic resonance

D
$$X-$$rays

Solution

Founer Transform Infrared(FIIR) spectromicroscopy could be used to detect contamination in finger print
Reason$$\rightarrow $$A substance can be identified because its chemical enviroment often shows up like a finger print in a spectrum.
$$\Rightarrow $$Infrared radiation
Question 10
1 / -0

The emission spectra of atoms in the gaseous phase do not show a continuous spread of wavelength from red to violet, rather they emit light only at specific wavelengths with dark spaces between them. Such spectra is/are called:

A
line spectra

B
atomic spectra

C
both $$(A)$$ and $$(B)$$

D
none of these

Solution

The emission spectra of atoms which do not show continuous spread of wavelength. but emit only at specific wavelength is known as line spectra.

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Re-Attempt Weekly Quiz Competition

Structure of Atom Test - 71

Result

Structure of Atom Test - 71

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SHARING IS CARING

Question 1

$$He^+$$

$$Li^{2+}$$

$$Be^{3+}$$

All of the these

Solution

Question 2

Sc

Ba

Br

I

Solution

Question 3

6

4

3

2

Solution

Question 4

$$2.07\overset{o}{A}$$

$$1.59\overset{o}{A}$$

$$0.53\overset{o}{A}$$

$$2.12\overset{o}{A}$$

Question 5

Independent of which orbit it is in

Negative

Positive

Increase with the quantum number $$n$$

Solution

Question 6

$$\frac{4\pi^2 mr^2}{nh}$$

$$\frac{nh}{4\pi^2 mr}$$

$$\frac{2\pi mr}{n^2h^2}$$

$$\frac{h}{2\pi mr}$$

Solution

Question 7

$$2$$

$$3$$

$$4$$

$$5$$

Solution

Question 8

$$1$$

$$2$$

$$3$$

none

Solution

Question 9

infrared radiation

ultraviolet light

nuclear magnetic resonance

$$X-$$rays

Solution

Question 10

line spectra

atomic spectra

both $$(A)$$ and $$(B)$$

none of these

Solution

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