MCQ Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 13

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:

0%
infrared radiation
0%
ultraviolet light
0%
nuclear magnetic resonance
0%
$X-$ rays

Explanation

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

An ionic atom equivalent to hydrogen atom has wavelength equal to

$\dfrac{1}{4}$ of the wavelengths of hydrogen lines. The ion will be

0%
$He^+$
0%
$Li^{++}$
0%
$Ne^{9+}$
0%
$Na^{10+}$

Explanation

$\text{By Rydberg's formula ,we have -}$

$\begin{array}{l}R_{H}=\text { Rydberg constant } \\\lambda=\text { wavelength } \\Z=\text { atomic num. }\end{array}$

$\begin{array}{l}\text { * For a particular member of spectral line }\\\text { we have - } \\\qquad \begin{aligned}\frac{1}{\lambda} &\alpha z^{2} \quad \Rightarrow\left[\lambda \alpha \frac{1}{z^{2}}\right] \\& \Rightarrow \lambda=\frac{k}{z^{2}} \text { (let) }-(1) \\\Rightarrow & \lambda_{H}=k\end{aligned}\end{array}$

$\begin{array}{l}\text { For any other ionic atom for which } \\\text { wavelength }=\frac{\lambda_{H}}{4}=\frac{k}{4}-(2) \\\text { from eq (1) and (2) } \\\text { we get } \frac{k}{z^{2}}=\frac{k}{4}\\\Rightarrow z^{2}=4 \\\Rightarrow z=2\quad\left(\mathrm{He}^{+}\right) \text {atom. } \\\text { Hence, option (A) is correct. }\end{array}$

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:

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

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,

$E_n=-R_H\left(\frac{1}{n^2}\right)$

$n^2=\frac{-R_H}{E_n}=\frac{-2.18\times 10^{-18}}{-1.5\times 1.6\times 10^{-19}}=9.08$

$n=3$

The gyromagnetic ratio of an electron in an

$H-$ atom, according to Bohr model, is:-

0%
Independent of which orbit it is in
0%
Negative
0%
Positive
0%
Increase with the quantum number $n$

Explanation

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

The frequency of the characterstic X ray of

$K_a$ line of metal targent 'M' is

$2500 cm^{-1}$ and the graph between

$\sqrt { v }$ Vs 'z' is as follows, then atomic number of M is?

0%
$49$
0%
$50$
0%
$51$
0%
$25$

Angular momentum in second Bohr orbit of H-atom is

$x$ . Then find out angular momentum in 1st excited state of

${Li}^{+2}$ ion:

0%
$3x$
0%
$9x$
0%
$\cfrac{x}{2}$
0%
$x$

Explanation

$\begin{array}{l} \text { we know, } \\ \text { Angular momentum }=\frac{n h}{2 \pi} \end{array}$

$\begin{array}{l} \text { Given, } \\ \text { For } n=2 \\ \text { Angular momentum = } x \\ \qquad x=\frac{2 h}{2 \pi} --(i) \end{array}$

$\text { for } 1^{\text {st }} \text { excited state means } n=2$

$\begin{aligned} & A \cdot M=\frac{2 h}{2 \pi} \quad-(11) \\ \therefore \quad &(i)=(i i) \end{aligned}$

$\text { Angular mementum }= x$

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:

0%
line spectra
0%
atomic spectra
0%
both $(A)$ and $(B)$
0%
none of these

If one were to apply Bohr's model to a particle of mass 'm' and charge 'q' moving in a plane under the influence of a magnetic field 'B', the energy of the charged particles in the

$n^{th}$ level will be:

0%
n $(\dfrac{hqB}{8\pi m})$
0%
n $(\dfrac{hqB}{4\pi m})$
0%
n $(\dfrac{hqB}{2\pi m})$
0%
n $(\dfrac{hqB}{\pi m})$

Explanation

According to given data

We know that,

$mvr=\dfrac{nh}{2\pi }$

Or,

$qvB=\dfrac{m{{v}^{2}}}{r}$

So we find,

$qB=\dfrac{mv}{r}$

$qB\left( \dfrac{nh}{2\pi mv} \right)=mv$

$\left( \dfrac{1}{2}m{{v}^{2}} \right)=\dfrac{nhqB}{4\pi m}$

$E=n\left( \dfrac{hqB}{4\pi m} \right)$

This is the required solution.

Assuming Bohr's model for

$Li^{++}$ atom, the first excitatio energy of ground state of

$Li^{++}$ atom is

0%
10.2 eV
0%
91.8 eV
0%
13.6 eV
0%
3.4 eV

Explanation

For

$Li^{++}$ model,

$2=3$

In first excitation of ground state,

$n_1=1$ and

$n_2=2$

In Bohr's model,

$\Delta E=(13.6ev)2^2\left[\dfrac{1}{n_1^2}-\dfrac{1}{n^2_2}\right]$

Here,

$\Delta E=(13.6)(3)^2\left[\dfrac{1}{1}-\dfrac{1}{4}\right]$

$\Delta E=(13.6)9\times \dfrac{3}{4}=91.8$ eV

So, first excitation energy for ground state in

$Li^{++}$ is

$91.8$ eV

Option B is correct.

1132942_1158681_ans_eea3cec6a454484b943cf447c060ab8a.jpg

The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of

$108.5nm$ . the ground state energy of an electron of this ion will be

0%
$3.4eV$
0%
$13.6eV$
0%
$54.4eV$
0%
$122.4eV$

Explanation

For third of Balmer series

$n_1=2,\ n_2=5$

$\therefore \dfrac{1}{\lambda}=RZ^2 \left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2} \right]$

gives

$Z^2=\dfrac{n_1^2 n_2^2}{(n_2^2-n_1^2) \lambda R}$

On putting values

$Z=2$

From

$E=-\dfrac{13.6Z^2}{n^2}=\dfrac{-13.6(2)^2}{(1)^2}=-54.4\ eV$

The wavelength of the first member of the Balmer series in hydrogen spectrum is

$x$

$\mathring{A}$ . Then the wavelength (in

$\mathring{A}$ ) of the first member of Lyman series in the same spectrum is

0%
$\cfrac{5}{27}x$
0%
$\cfrac{4}{3}x$
0%
$\cfrac{27}{5}x$
0%
$\cfrac{5}{36}x$

Explanation

$\begin{array}{l} \text { Balmer series : transition take place from } \\ n=2 \text { to } n=3,4,5, \text { so on } \\ \text { for first member of balmer } \\ n_{1}=2 \text { to } n_{2}=3 \end{array}$

$\lambda=x \dot {A}$

$\begin{array}{l} \frac{1}{\lambda}=R z^{2}\left(\frac{1}{n_{1} 2}-\frac{1}{n_{2}^{2}}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{1}{4}-\frac{1}{9}\right) \\ \frac{1}{x}=R z^{2}\left(\frac{5}{36}\right)-(1) \end{array}$

$\begin{array}{l} \text { For first member of Lyman } \\ \qquad \begin{aligned} n_{1} &=3 \quad n_{2}=2 \\ \frac{1}{\lambda} &=R_{2}^{2}\left(\frac{1}{1}-\frac{1}{4}\right) \\ \frac{1}{\lambda} &=R z^{2}\left(\frac{3}{4}\right) \end{aligned} \end{array}$

$\begin{array}{l} \text { divide (i) } \div \text { (ii) } \\ \qquad \begin{array}{l} \frac{ \lambda }{x}=\left(\frac{5}{36}\right)\left(\frac{4}{3}\right) \\ \lambda=\left(\frac{5}{27}\right) x \end{array} \end{array}$

In a hypothetical system a particle of mass

$m$ and charge

$-3q$ is moving around a very heavy particle having charge

$q$ . Assuming Bohr's model to be true to this system, the orbital velocity of mass

$m$ when it is nearest to the heavy particle is

0%
$\cfrac{3{q}^{2}}{2{\epsilon}_{0}h}$
0%
$\cfrac{3{q}^{2}}{4{\epsilon}_{0}h}$
0%
$\cfrac{3{q}^{}}{2{\epsilon}_{0}h}$
0%
$\cfrac{3{q}^{}}{4{\epsilon}_{0}h}$

The mass of two moles of nitrogen atom is_____

0%
$28\space g$
0%
$14\space g$
0%
$7\space g$
0%
$20\space g$

If the boiling point of ethanol (molecular weight = 46 ) is

$78^o$ C, the boiling point of diethyl ether (molecular weight = 74) is

0%
$100^o$ C
0%
$78^o$ C
0%
$86^o$ C
0%
$34^o$ C

Find an expression for de Broglie wavelength for an electron in the

$n^{th}$ orbit of the Bohr model of the hydrogen atom and hence find its value when the electron is in n = 4 level.
Take radius of nth orbital of Bohr's model of H-atom;

$r_n = (53 \ n^2)$ pm

0%
1.33 nm
0%
1.20 nm
0%
1.50 nm
0%
None

Which principle/ rule limits the maximum number of electrons in orbital to two?

0%
Aufbau principle
0%
Pauli's exclusion principle
0%
Hund's rule of maximum multiplicity
0%
Heisenberg's uncertainty principle

Which of the following are isotopes?
(i) Atom, whose nucleus contains 20p+15n
(ii) Atom whose nucleus contains 20p+17n
(iii) Atom, whose nucleus contains 18p+22n
(iv) Atom, whose nucleus contains 18p+21n

0%
(i) and (iii)
0%
(i) and (ii)
0%
(ii) and (iii)
0%
(ii) and (iv)

Which of the following elements is non- radioactive?

0%
plutonium
0%
Zirconium
0%
Thorium
0%
Uranium

Calculate minimum number of electrons in

$Fe$ atom having

$(n+l) \ge 4,\ m=-1, \ s=+\frac { 1 }{ 2 }$ .

0%
1
0%
2
0%
3
0%
4

An electron jumps from the

$4^{th}$ orbit to the

$2^{nd}$ orbit of hydrogen atom

$(R=10^{5}\ Cm^{-1})$ . Frequency is in

$Hz$ of emitted radiation will be

0%
$\dfrac {3\ \times 10^{5}}{16}$
0%
$\dfrac {3\ \times 10^{15}}{16}$
0%
$\dfrac {9\ \times 10^{15}}{16}$
0%
$\dfrac {9\ \times 10^{5}}{16}$

Explanation

$\begin{array}{l}\text { Electron jumps from } 4^{\text {th }} \text { orbit to } \\2^{\text {nd }} \text { orbit. } \\\text { Let wavelength be "x" then } \\\qquad \frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\end{array}$

$\text { Here, } \begin{aligned}& R=10^{5} \mathrm{~cm}^{-1} \\& z=1\\& n_{1}=2, n_{2}=4 \\\frac{1}{\lambda} &=10^{5}\left[\frac{1}{2^{2}}\frac{1}{4^{2}}\right] \\\frac{1}{\lambda}&=10^{5}\left[\frac{1}{4}-\frac{1}{16}\right] \\\frac{1}{\lambda} &=10^{5}\left[\frac{3}{16}\right] \\\frac{1}{\lambda} &=\frac{3 \times 10^{5}}{16}-(1)\end{aligned}$

$\begin{array}{l}\text { Frequency is } \frac{c}{\lambda} \\c=3 \times 10^{8} \mathrm{~m}=3 \times 10^{10} \mathrm{~cm} \\\text { From frequency, } v=3 \times 10^{10} \times \dfrac{3 \times 10^{5}}{16}\\\qquad v=\dfrac{9 \times 10^{15}}{16}\end{array}$

Which of the following radiations has largest energy ?

0%
$\lambda = 30$ nm
0%
$\lambda = 300$ pm
0%
$v = 3\times 10^{12}$ s $^{-1}$
0%
$v = 3\times 10^{10}$ s $^{-1}$

According to Bohr's theory of the hydrogen atom, the speed

$v_n$ of the electron in a stationary orbit is related to the principal quantum number n as (C is a constant).

0%
$v_n=C/n^2$
0%
$v_n=C/n$
0%
$v_n=C\times n$
0%
$v_n=C\times n^2$

The average life of an excited state of hydrogen atom is of the order

${ 10 }^{ -8 }s$ . The number of revolution made by an electron when it is in state n= 2 and before it suffers a transition to state n = 1 are

0%
${ 8.23\times 10 }^{ 6 }$
0%
${ 2.82\times 10 }^{ 6 }$
0%
${ 22.8\times 10 }^{ 6 }$
0%
${ 2.28\times 10 }^{ 6 }$

Atomic number of

$N$ is

$7$ . The atomic number of 3rd member of nitrogen family is

0%
23
0%
15
0%
33
0%
43

Explanation

Period	I	II	III	IV	V
Element	N	P	$As$	Sb	Bi
Atomic no.	7	15	$33$	51	83

The atomic number of 3rd member of nitrogen family is

$33.$

In a hypothetical system a particle of mass m and charge -3 q is moving around a very heavy particle having charge q. Assuming Bohr's model to be true to this system, the orbital velocity of mass m when it is nearest to heavy particle is:

0%
$\frac{3q^2}{2\varepsilon _0h}$
0%
$\frac{3q^2}{4\varepsilon _0h}$
0%
$\frac{3q}{2\varepsilon _0h}$
0%
$\frac{3q}{4\varepsilon _0h}$

${ M }^{ 3+ }$ has electronic configuration as

$[Ar]{ 3d }^{ 10 }{ 4s }^{ 2 }$ . Hence it lies in:

0%
s-block
0%
p- block
0%
d-block
0%
f-block

Which of the following is inversely proportional to the square of principal quantum number in Bohr's atomic model?

0%
Total energy of the electron in $n^{th}$ orbit.
0%
Radius of the $n^{th}$ orbit.
0%
Frequency of revolution in $n^{th}$ orbit.
0%
Time period of revolution in $n^{th}$ orbit.

In an atom having

$2K, 8L, 8M$ and

$2N$ electrons, the number of electrons with

$m=0$ ;

$s=+\dfrac{1}{2}$ are?

0%
$6$
0%
$2$
0%
$8$
0%
$16$

Suppose that in any Bohr atom (or) ion orbits are only in even numbers like

$2,4,6,....$ the maximum wavelength of radiation emitted in the visible region of hydrogen spectrum should be ______________.

0%
$\dfrac{4}{R}$
0%
$\dfrac{R}{4}$
0%
$\dfrac{36}{5R}$
0%
$\dfrac{16}{3R}$

In a hydrogen like sample electron is in

$2nd$ excited state, the binding energy of

$4th$ state of this sample is

$-13.6\ eV$ , then incorrect statement is :

0%
Atomic number of element is $4$ .
0%
$3$ different types of spectral line will be observed if electrons make transition upto ground state from the $2nd$ excited state.
0%
A $25\ eV$ photon can set free the electron from the $2nd$ excited state of this sample.
0%
$2nd$ line of Balmer series of this sample has same energy value as $1st$ excitation energy of $H-$ atoms.

Explanation

As we know,

$E=-13.6 \frac{Z^2}{n^2}.$

For n=5(4th excited state) Z=5. For Z=5 E and n =3(2nd excited state) E is

$13.6* \frac{25}{9}=37.77$

Sinece 37.7 > 25 therefore the electron cant be set free .

As all options match except D.

Hence , option D is correct .

What is the maximum number of electrons that can be accommodated in the outermost orbit?

0%
2
0%
8
0%
3
0%
18

The speed of sound in air is

$v$ . Both the source and observer are moving towards each other with equal speed

$u$ . The speed of wind is

$w$ from source to observer. Then, the ratio

$(\cfrac{f}{{f}_{0}})$ of the apparent frequency to the actual frequency is given by

0%
$\cfrac{v+u}{v-u}$
0%
$\cfrac{v+w+u}{v+w-u}$
0%
$\cfrac{v+w+u}{v-w-u}$
0%
$\cfrac{v-w+u}{v-w-u}$

Explanation

$\begin{array}{l} f=\left( { \dfrac { { v+u+w } }{ { v-u+w } } } \right) { f_{ 0 } } \\ \Rightarrow \dfrac { f }{ { { f_{ 0 } } } } =\left( { \dfrac { { v+u+w } }{ { v-u+w } } } \right) \\ Hence, \\ option\, \, C\, \, is\, \, correct\, \, answer. \end{array}$

Which of the following orbitals has the lowest energy?

0%
$4d$
0%
$4f$
0%
$5s$
0%
$5p$

Explanation

The energy of the orbitals are calculated using

$n+l$ values of orbitals.

$4d=4+2=6$

$4f=4+3=7$

$5s=5+0=5$

$5p=5+1=6$

$5s$ lowest energy.

Hence, option C is correct.

According to Moseley's law, the ratio of the slope of graph between

$\sqrt { f }$ and

$Z$ for

${ k }_{ \beta }$ and

${ k }_{ \alpha }$ is

0%
$\sqrt { \dfrac { 32 }{ 27 } }$
0%
$\sqrt { \dfrac { 27 }{ 32 } }$
0%
$\sqrt { \dfrac { 5 }{ 56 } }$
0%
$\sqrt { \dfrac { 36 }{ 5 } }$

The ratio of masses of an element A attached with the fixed mass of another element B in their two compounds is 2:If 48 mass of A is attached with 24 mass of B in first compound then what mass of A is attached with the 12 mass of b in the second compound?

0%
72
0%
36
0%
24
0%
12

The electric potential between a proton and an electron is given by

$V=VIn\dfrac { r }{ { r }_{ 0 } }$ , where

${ r }_{ 0 }$ is a constant. Assuming Bohr's model to be applicable,write variation of

${ r }_{ n}$ with

$n,\ n$ being the principal quantum number.

0%
${ r }_{ n }\propto n$
0%
${ r }_{ n }\propto \dfrac { 1 }{ n }$
0%
${ r }_{ n }\propto { n }^{ 2 }$
0%
${ r }_{ n }\propto \dfrac { 1 }{ { n }^{ 2 } }$

For an atom with atomic number 14, the number of orbits and orbitals in which electrons are present is respectively:

0%
$3,6$
0%
$6,3$
0%
$7,3$
0%
$3,8$

One Bohr magneton is equal to ____________

${ AM }^{ 2 }$ .

0%
$4.8\times { 10 }^{ -10 }eSu$
0%
$1.6\times { 10 }^{ -19 }eSu$
0%
$4.8\times { 10 }^{ -10 }\ eSu\ \& 1.6\times { 1 0}^{ -19 } \ eSu$
0%
$9.27\times { 10 }^{ -24 }$

The de Broglie wavelength of an electron in

$n^{th}$ Bohr orbit of radius

$a_n(a_0$ being the first orbit radius

$)$ :

0%
$2\pi a_0$
0%
$2\pi n^2a_0$
0%
$2\pi na_0$
0%
$\dfrac{2\pi a_0}{n}$

The ground state energy of hydrogen atom is

$-13.6$ eV. The energy of second excited state

$He^+ ion$ in eV is:

0%
$-6.04$
0%
$-27.2$
0%
$-54.4$
0%
$-3.4$

Explanation

$(E)_n^{th}=(E_{GND})_H^.\dfrac { Z^2 }{ n^2 }$

$E_{3rd}(He^+)=(-13.6eV)\times\dfrac { 2^2}{ 3^2 } =-6.04eV$

After the filling of np-orbitals next orbital filled will be:

0%
ns
0%
nd
0%
(n-1)d
0%
(n+1)s

The number of the spectral lines obtained when an electron is provided with a photon of the energy of

$11\ eV$ (for H atom) is/are:

0%
$1$
0%
$2$
0%
$0$
0%
$6$

Imagine an atom made of a proton and a hypothetical particle of double the mass as that of an electron but the same charge. Apply Bohr theory to consider transitions of the hypothetical particle to the ground state . Then, the longest wavelength (in terms of Rydberge constant for hydrogen atom ) is

0%
$\dfrac { 1 } { 2 R }$
0%
$\dfrac { 5} { 3 R }$
0%
$\dfrac { 1 } { 3 R }$
0%
$\dfrac { 2 } { 3 R }$

Which is correct statement about proton?

0%
Proton is nucleus of deuterium.
0%
Proton is ionized oxygen molecule.
0%
Proton is ionized hydrogen atom.
0%
None of these

Explanation

When a hydrogen atom loses an electron, it becomes an ion having only one proton and no electrons. So, we can say that a proton is an ionised hydrogen atom.

$H \xrightarrow{Loses\ electron} H^+ (proton)$

FUNDAMENTAL PARTICLES
Nucleus was discovered in

0%
discharge of electricity through gases
0%
the bormbarding experiment Wilson
0%
$\alpha$ -ray scattering experiment
0%
Wilson cloud chamber

The time taken by the electron in one complete revolution in the

$n^{th}$ Bohr's orbit of the hydrogen atom is

0%
Inversely proportional to $n^2$
0%
Directly proportional to $n^3$
0%
Directly proportional to $\dfrac{h}{2\pi}$
0%
Inversely proportional to $\dfrac{n}{h}$

Which of the following elements has two shells, both of which are completely filled with electrons?

0%
Hydrogen (H)
0%
Boron (B)
0%
Helium (He)
0%
Neon (Ne)

The hydrogen atom in ground state is excited by a monochromatic radiation of

$\lambda=975 A^{\circ}$ . Number of spectral lines in the resulting spectrum emitted will be

0%
3
0%
2
0%
6
0%
10

Consider the following given species and find out, for which Bohr's atomic model is not applicable?

0%
$1^{H^{2}}$
0%
$He^{ + }$
0%
$H^{+}$
0%
$\mathrm{Li}^{2+}$

Which of the following gas is mono-atomic?

0%
Chlorine
0%
Helium
0%
Hydrogen
0%
Oxygen

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 13 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.