Statement I : $$\displaystyle ^{ 12 }{ C }\ $$ is an isotope of $$\displaystyle ^{ 14 }{ C }\ $$. Statement II: The nuclei of both atoms have the same number of neutrons.

true, true, correct explanation

An element $$X$$ combines with oxygen to form compounds $$P$$ and $$Q$$. If the ratio of the valency of element $$X$$ in $$P$$ to element $$X$$ in $$Q$$ is $$3 : 5$$ respectively, what could be the probable compounds $$P$$ and $$Q$$?

$$\begin{matrix} P & Q \\ { N }_{ 2 }{ O }_{ 3 } & { N }_{ 2 }{ O }_{ 5 } \end{matrix}$$

$$\begin{matrix} P & Q \\ CO & C{ O }_{ 2 } \end{matrix}$$

$$\begin{matrix} P & Q \\ { H }_{ 2 }O & { H }_{ 2 }{ O }_{ 2 } \end{matrix}$$

$$\begin{matrix} P & Q \\ C{ H }_{ 4 } & { C }_{ 3 }{ H }_{ 6 } \end{matrix}$$

MCQ Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 12

To explain his theory, Bohr-used

0%
Conservation of linear momentum
0%
Conservation of angular momentum
0%
Conservation of quantum frequency
0%
Conservation of energy

In Bohr's model of hydrogen atom, which of the following pairs of quantities are quantized

0%
Energy and linear momentum
0%
Linear and angular momentum
0%
Energy and angular momentum
0%
None of these

The main particle of the outer part of an atom is

0%
Proton
0%
Neutron
0%
Proton and Neutron
0%
Electron

Which of the following is quantised according to Bohr's theory of hydrogen atom

0%
Linear momentum of electron
0%
Angular momentum of electron
0%
Linear velocity of electron
0%
Angular velocity of electron

The total number of atoms per unit cell in bcc is:

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

Explanation

B C C

$BCC$ , there is one atom at the centre and one - eight fraction of atom at every corner.

So,

1 + \frac{1}{8} \times 8 = 2

$1 + \dfrac{1}{8} \times 8 = 2$ atom.

The boher model of atoms

0%
Assume that the angular momentum of electrons is quantized
0%
Uses Einstein's photo electric equation
0%
Predicts continuous emission spectra for atoms
0%
Predicts the same emission spectra for all types of atoms

In hydrogen atom which quantity is integral multiple of

\frac{h}{2 π}

$\dfrac{h}{2\pi}$

0%
Angular momentum
0%
Angular velocity
0%
Angular acceleration
0%
Momentum

When a hydrogen atom is raised from ground to excited state.

0%
The P.E increases and K.E decreases
0%
The P.E decreases and K.E increases
0%
Both P.E and K.E increase
0%
Both P.E and K.E decreases

Explanation

Potential energy of hydrogen atom is given by

P . E = \frac{- k Z e}{r}

$P.E=\dfrac{-kZe}{r}$

As atom is excited,it's radius increases. But due to negative sign or reverse sign. Its actual potential energy increases.

And kinetic energy is given by

K . E = \frac{k Z e}{2 r}

$K.E=\dfrac{kZe}{2r}$

As Tom is excited, it's radius increases this resulting in decrease of kinetic energy.

The PE increases and KE decreases.

n = 3, l = 0, m = 0

$n = 3 , l = 0 , m = 0$ then atomic number is

0%
$12 , 13$
0%
$13 , 14$
0%
$10 , 11$
0%
$11 , 12$

Explanation

n = 2, l = 0

$n = 2 , l = 0$ means last shell is

3 s

$3s$
E.C will be

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1 - 2}

$1s^{2} 2s^{2} 2p^{6} 3s^{1 - 2}$
Atomic no. is

11

$11$ or

12

$12$ .

In the ground state, an element has

13

$13$ electrons in its M-shell. The element is

0%
Manganese
0%
Chromium
0%
Nickel
0%
Iron

Explanation

13 e^{-}

$13e^{-}$ in M

(3 r d)

$(3rd)$ shell means

3 s^{2} 3 p^{6} 3 d^{5}

$3s^{2} 3p^{6} 3d^{5}$ . hence complete configuration will be

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{5}

$1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{1} 3d^{5}$
i.e., total

e^{-} = 24

$e^{-} = 24$ . Hence the element is chromium.

The maximum number of electrons in an orbit with

l = 2, n = 3

$l = 2 , n = 3$ is

0%
$2$
0%
$6$
0%
$12$
0%
$10$

Explanation

n = 3, l = 2

$n = 3 , l = 2$ means

3 d

$3d$ -subshell. Maximum number of electrons present in it

= 10

$= 10$

Any p-orbital can accommodate upto

0%
four electrons
0%
two electrons with parallel spin
0%
six electrons
0%
two electrons with anti-parallel spin.

Naturally occurring B consists of two isotopes, whose atomic weights are 10.01 and 11.The atomic weight of natural boron is 10.The % of each isotope in natural boron is :

0%
% of isotope of mass 10.01 = 20, % of isotope of mass 11.01 = 80
0%
% of isotope of mass 10.01 = 30, % of isotope of mass 11.01 = 70
0%
% of isotope of mass 10.01 = 50, % of isotope of mass 11.01 = 50
0%
none of these

Explanation

Average atomic weight =

\frac{relative atomic mass of type 1 X % abundance of type 1 + relative mass of type 2 X % abundance of type 2}{100}

$\cfrac{ \text {relative atomic mass of type 1 X % abundance of type 1} + \text {relative mass of type 2 X % abundance of type 2 }}{100}$

Let % of isotope of mass 10.01 be

a

$a$ .

10.81 = \frac{10.01 \times a + 11.01 (100 - a)}{100}

$10.81\, =\, \displaystyle \frac {10.01\, \times\, a\, +\, 11.01(100\, -\, a)}{100}$

a = 20

$a = 20$

% of isotope of mass 10.01 = 20

% of isotope of mass 11.01 = 80

In what ratio Cl

_{17}^{37}

$_{17}^{37}$ and Cl

_{17}^{35}

$_{17}^{35}$ be present so as to obtain Cl

_{17}^{35.5}

$_{17}^{35.5}$ ?

0%
1 : 2
0%
1 : 1
0%
1 : 3
0%
3 : 1

Explanation

Let

x

$x$ % of

C l

$Cl$

^{37}

$^{37}$ and

(100 - x)

$(100 - x)$ % of

C l

$Cl$

^{35}

$^{35}$ be present in

C l

$Cl$

_{17}^{35.5}

$_{17}^{35.5}$ .

35.5 =

$35.5 =$

\frac{x \times 37 + (100 - x) \times 35}{100}

$\displaystyle\frac{x \times 37 + (100 - x)\times 35}{100}$

\Rightarrow x = 25

$\Rightarrow x = 25$

∴ %

$\therefore \%$ of

C l

$Cl$

^{37}

$^{37}$

= 25

$= 25$

∴ %

$\therefore \ \%$ of

C l

$Cl$

^{35}

$^{35}$

= 75

$= 75$

∴

$\therefore$ Ratio is

= 25 : 75 = 1 : 3

$=25:75=1 : 3$ .

According to Bohr's theory of hydrogen-atom, for the electron in the

n^{t h}

$n^{th}$ permissible orbit

0%
$\mbox{Linear Momentum } \propto \displaystyle\frac{1}{n}$
0%
$\mbox{Radius of orbit } \propto n$
0%
$\mbox{Kinetic Energy } \propto \displaystyle\frac{1}{n^2}$
0%
$\mbox{Angular Momentum } \propto n$

Explanation

Angular Momentum Quantisation:

m v r = \frac{n h}{2 π}

$mvr = \cfrac{nh}{2 \pi}$ ------------(1)

Centripetal force:

\frac{k (e) e}{r^{2}} = \frac{m v^{2}}{r}

$\cfrac{k(e)e}{r^2} = \cfrac{mv^2}{r}$ ----------------(2)

v^{2} r = \frac{k e^{2}}{m}

$v^2 r = \cfrac{ke^2}{m}$

Substituting (1) in (2)

v = \frac{2 π k e^{2}}{n h}

$v= \cfrac{2 \pi ke^2}{n h}$

Energy:

E = 13.6 \frac{Z^{2}}{n^{2}}

$E = 13.6 \cfrac{Z^2}{n^2}$ -----------(3)

Linear Momentum:

m v r = n h / 2 π

$mvr = nh/2 \pi$

m v = \frac{n h}{2 π r}

$mv = \cfrac{nh}{2 \pi r}$ ---------------(4)

From the concept attached:

r \propto n^{2}

$r \propto n^2$ -----------------(5)

Hence,

m v \propto \frac{1}{n}

$mv \propto \cfrac{1}{n}$ by (6)
and

U = 2 K E

$U = 2KE$

E \propto \frac{1}{n^{2}}

$E \propto \cfrac{1}{n^2}$ (concept attached)

How are isotopes of the same element different?

0%
They will have different locations on the periodic table.
0%
They will have different numbers of electrons.
0%
They will have undergo different chemical reactions.
0%
They have different numbers of protons.
0%
They have different numbers of neutrons.

If Aufbau rule is not obeyed and the electronic filling occurs orbit by orbit till saturation is reached (for a given principal quantum number

n

$n$ , electrons are filled in the increasing order of azimuthal quantum number

l

$l$ ), then the percentage change in sum of all

(n + l)

$(n + l)$ values for unpaired electrons in an atom of iron is

10 x

$10x$ . What is the value of

x

$x$ ?

0%
$5$
0%
$6$
0%
$7$
0%
$3$

Explanation

When

n + l

$n+l$ rule is obeyed, the configuration is

^{26} F e = 1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 4 s^{2}, 3 d^{6} .

$^{26}Fe= 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{2}, 3d^{6}.$

Since,

(n + l)

$(n+l)$ for each unpaired electron is

5

$5$ , so for

4

$4$ unpaired electrons in

^{26} F e

$^{26}Fe$ ,

(n + l)

$(n+l)$ is

5 \times 4 = 20

$5\times 4= 20$ .

When

n + l

$n+l$ rule is not obeyed, the configuration is

^{26} F e = 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{8} .

$^{26}Fe= 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{8}.$

Since,

(n + l)

$(n+l)$ for each unpaired electron is

5

$5$ , so for

2

$2$ unpaired electrons in

^{26} F e

$^{26}Fe$ ,

(n + l)

$(n+l)$ is

5 \times 2 = 10

$5\times 2= 10$ .

Now,
% change

= \frac{20 - 10}{20} \times 100 = 50 = 10 x

$= \dfrac{20-10}{20} \times 100 = 50 = 10x$ .

∴ x = 5.

$\therefore x=5.$

What principle is violated here?

0%
Laws of Motion
0%
Energy conservation
0%
Nothing is violated
0%
Cannot be decided

Explanation

Laws of Motion:
Newtons second law:

F = m a

$F = ma$
Without any force, the atom cannot accelerate.
Conservation of Momentum:
The linear momentum of the atom can't change without any external force acting on it.

In Bohr's model of an atom :

0%
Electrons revolve around the nucleus in circular orbits.
0%
These circular orbits or shells are stationary.
0%
Each shell in associated with a definite amount of energy.
0%
All shells have equal energy level.

Maximum number of electrons in a subshell of an atom is determined by the following:

0%
$\text{2l+2}$
0%
$\text{4l-1}$
0%
$2n^2$
0%
$\text {4l + 2}$

Explanation

Hint

Quantum number: It is defined as the set of numbers which describes the position and energy of electrons in an atom. There are four quantum numbers: principal, azimuthal, magnetic and spin quantum numbers.

Explanation

Step 1: Determine the maximum number of subshells

The principle quantum number ( $n$ ) describe the distance between the nucleus and the electrons.
The azimuthal quantum number ( $l$ ) is given by ( $n - l$ )

We know that the maximum number of subshells is equal to is $\left( 2l+1 \right)$ .

Step 2: Determine the maximum number of electrons in a subshell
Maximum electrons a subshell can accommodate is $2$

Therefore, the total number of electrons in a subshell is

$2\left( 2l+1 \right)=\left( 4l+2 \right)$

Thus, maximum number of electrons in a subshell is $( 4l+2)$

Final answer

The correct answer is option (D).

Many elements have fractional atomic mass. This is because of :

0%
they have isotopes
0%
their isotopes have non-integral masses
0%
their isotopes have different masses
0%
neutrons, protons and electrons combine to give fractional masses

Explanation

Several elements have isotopes with different atomic masses. Hence, many elements have fractional atomic masses.

Thus, options A and C are correct.

For example, chlorine has two isotopes with atomic masses

35

$35$ and

37

$37$ with a natural abundance ratio of

3 : 1

$3:1$ .

\frac{35 \times 3 + 37 \times 1}{4} = 35.5

$\dfrac{35\times 3 + 37\times 1}{4} = 35.5$

The weighted average atomic mass is

35.5

$35.5$ .

For the hydrogen atom, the energy of the electron is defined by the factor

E_{n} = - 13.58 / n^{2} e V

$E_n=-13.58/n^2 eV$ . Here n is positive integer. The minimum quantity of energy which it can absorb in its primitive stage is:

0%
$1.00 eV$
0%
$3.39 eV$
0%
$6.79 eV$
0%
$10.19 eV$

Explanation

E_{n} = \frac{- 13.58}{n^{2}}

$E_n=\dfrac {-13.58}{n^2}$

In primitive stage at

n = 1, E_{1} = - 13.58 e V

$n=1, E_1=-13.58 eV$

Minimum energy in excited stage,

n = 2, E_{2} = - 3.395 e V

$n=2,\quad E_2=-3.395 eV$

Energy absorbed

= - 3.395 + 13.58 = 10.19 e V

$=-3.395+13.58=10.19eV$

Hence, the correct option is D

The total number of electrons that could be held in a sub-level with azimuthal quantum number = 2 is :

0%
$2$
0%
$6$
0%
$8$
0%
$10$

Explanation

The azimutual quantum number,

^{'} l^{'} = 2

$'l' = 2$ corresponds to 'd-subshell'.
A 'd-subshell' has 5 orbitals, and in each orbital there a maximum of 2 electrons can be filled.
The total number of electrons present in all 5 orbitals of d-subshell

= 5 \times 2 = 10

$= 5 \times 2 = 10$

The maximum number of electrons in a shell with the principal quantum number = 4 is :

0%
$2$
0%
$10$
0%
$16$
0%
$32$

Explanation

Formula to determine the number of electrons present in any shell is as follows :
Let the shell number is "n", then number of electrons present in n shell =

2 n^{2}

$2n^{2}$ .
Hence, as per formula the number of electrons iin 4th shell =

2 \times 4^{2} = 32

$2 \times 4^{2} = 32$ .

If the principle quantum number n = 6, the correct sequence of filling of electron will be :

0%
$ns\rightarrow$ $(n - 2)f$ $\rightarrow$ np $\rightarrow (n - 1)$
0%
$ns\rightarrow$ $(n - 1)f$ $\rightarrow$ $np$ $\rightarrow (n - 2)$
0%
$ns\rightarrow np\rightarrow(n - 1)d\rightarrow(n - 2)f$
0%
$ns\rightarrow (n - 2)f \rightarrow (n - 1)d \rightarrow np$

Explanation

If the principle quantum number n = 6, the correct sequence of filling of electron will be:

n s \to (n - 2) f \to (n - 1) d \to

$ns\rightarrow(n - 2)f\rightarrow(n - 1)d\rightarrow$ np.

Here,

n = 6

$n=6$ . Hence, the correct sequence of filling of electron will be

6 s \to 4 f \to 5 d \to

$6s\rightarrow4f\rightarrow5d\rightarrow$

6 p

$6p$ .

Hence, option D is correct.

Statement I :

^{12} C

$\displaystyle ^{ 12 }{ C }\$ is an isotope of

^{14} C

$\displaystyle ^{ 14 }{ C }\$ .
Statement II: The nuclei of both atoms have the same number of neutrons.

0%
true, false
0%
false, true
0%
false, false
0%
true, true, correct explanation

Explanation

^{12} C

$^{12}C$ is an isotope of

^{14} C

$^{14}C$ .

As isotopes are variants of a particular chemical element that differ in neutron number but have the same number of protons.

^{12} C

$^{12}C$ has

12 - 6 = 6

$12 - 6 = 6$ neutrons.

^{14}

$^{14}$ has

14 - 6 = 8

$14 - 6 = 8$ neutrons.

Hence, the statement

I

$I$ is true but the statement

I I

$II$ is false.

If the principal quantum number of a shell is 2, what types of orbitals will be present?

0%
$s$
0%
$s$ and $p$
0%
$s,\ p$ and $d$
0%
$s,\ p,\ d$ and $f$

Explanation

For a shell with the principal quantum number "

n

$n$ ", the types of orbitals are determined by the value of the magnetic quantum number "

l

$l$ ".

For a shell with the principal quantum number "

n

$n$ ", the '

l

$l$ ' values ranges from

0

$0$ to

n - 1

$n-1$ .

When

n = 2

$n = 2$ then '

l

$l$ ' has values

= 0, 1

$= 0 ,\ 1$ .

l = 0

$l = 0$ means type of the orbitals is s-orbital.

l = 1

$l = 1$ means type of orbitals is p-orbital.
Hence, there are

s

$s$ and

p

$p$ orbitals for a shell with the principal quantum number "

n

$n$ ".

Designations of some orbitals are given below. Arrange the orbitals with possible designations in the order in which they are filled with electrons

6 s, 8 p, 7 s, 4 d, 2 p, 3 d, 3 f, 4 f

$6s,\ 8p,\ 7s,\ 4d,\ 2p,\ 3d,\ 3f,\ 4f$ .

0%
$2p < 3d < 4d < 4p < 6s < 8p < 7s<4f$
0%
$2p < 3d < 4p < 4d < 6s < 4f < 7s < 8p$
0%
$3d < 2p < 4p < 4d < 6s < 4f < 7s < 8p$
0%
$2p < 3d < 4p < 4d < 4f < 6s < 7s < 8p$

Explanation

The orbitals with possible designations in the order in which they are filled with electrons is

2 p < 3 d < 4 p < 4 d < 6 s < 4 f < 7 s < 8 p

$2p < 3d < 4p < 4d < 6s < 4f < 7s < 8p$ .

Which letter orbital corresponds to

l

$l$ = 2?

0%
S
0%
p
0%
d
0%
f
0%
n

Explanation

l = 0

$l =0$ , means s

l = 1

$l = 1$ means p

l = 2

$l = 2$ means d

l = 3

$l = 3$ means f

Which color of light has the highest energy?

0%
Violet
0%
Green
0%
Yellow
0%
Orange
0%
Red

Explanation

Violet has highest frequency.

E = h v

$E = hv$

v = f r e q u e n c y,

$v = frequency,$
Violet has highest energy.

An element

X

$X$ combines with oxygen to form compounds

P

$P$ and

Q

$Q$ . If the ratio of the valency of element

X

$X$ in

P

$P$ to element

X

$X$ in

Q

$Q$ is

3 : 5

$3 : 5$ respectively, what could be the probable compounds

P

$P$ and

Q

$Q$ ?

0%
$\begin{matrix} P & Q \\ CO & C{ O }_{ 2 } \end{matrix}$
0%
$\begin{matrix} P & Q \\ { N }_{ 2 }{ O }_{ 3 } & { N }_{ 2 }{ O }_{ 5 } \end{matrix}$
0%
$\begin{matrix} P & Q \\ { H }_{ 2 }O & { H }_{ 2 }{ O }_{ 2 } \end{matrix}$
0%
$\begin{matrix} P & Q \\ C{ H }_{ 4 } & { C }_{ 3 }{ H }_{ 6 } \end{matrix}$

Explanation

Let

P = X_{2} O_{a}

$P = X_2O_a$ where a is the valency of X in P.

and let

Q = X_{2} O_{b}

$Q = X_2O_b$ where b is the valency of X in Q

According to the question:

The ratio of valency of X in P to element X in Q is

3 : 5

$3 : 5$ respectively.

Thus,

\frac{a}{b} = \frac{3}{5}

$\dfrac{a}{b} = \dfrac{3}{5}$

Therefore,

a = 3

$a = 3$ and

b = 5

$b = 5$

∴

$\therefore$ compound

P = X_{2} O_{3}

$P = X_2 O_3$

and compound

Q = X_{2} O_{5}

$Q = X_2 O_5$

According to the options, option (B) is satisfied, So,

P = N_{2} O_{3}

$P = N_2O_3$ and

Q = N_{2} O_{5}

$Q = N_2 O_5$

Hence, the correct option is (B)

The ratio of the energy of photon of

2000 A^{0}

$2000 A^0$ wavelength to that of

4000 A^{0}

$4000 A^0$ wavelength is:

0%
1 : 4
0%
4 : 1
0%
1 : 2
0%
2 : 1

Explanation

E = \frac{h c}{λ} \Rightarrow E \propto \frac{1}{λ}

$E=\cfrac { hc }{ \lambda } \Rightarrow E\propto \cfrac { 1 }{ \lambda }$

∴ \frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{4000}{2000} = 2 : 1

$\therefore \cfrac { { E }_{ 1 } }{ { E }_{ 2 } } =\cfrac { { \lambda }_{ 2 } }{ { \lambda }_{ 1 } } =\cfrac { 4000 }{ 2000 } =2:1$

Hence, option

D

$D$ is correct.

Two trials of a reaction were run in a lab where absorbance data was collected over time.
In trial

2

$2$ , the absorbance values for the reaction for each unit of time were higher than they were for trial

1

$1$ .
Also, the rate of trial

2

$2$ was higher than that of trial

1

$1$ .
Which of the following is most likely the case?

0%
The concentration of one or more reactants was increased in trial $2$ .
0%
The concentration of the reactants in trial $1$ was higher than that in trial $2$ .
0%
The concentration of reactants was decreased for trial $2$ .
0%
Something interfered with the reaction.

Explanation

Absorbance

= ε \times c o n c e n t r a t i o n \times p a t h l e n g t h \dots [B e e r l a w]

$=\varepsilon \times concentration \times path length\dots [Beer \quad law]$

∴

$\therefore$ Absorbance

\propto

$\propto$ concentration.

Also rate of reaction

\propto

$\propto$ concentration of reactant

∴

$\therefore$ It is possible that the concentration of one or more reactant was increased in trial

2

$2$ , giving higher values in trial

2

$2$ .

Which of the following statements is TRUE of matter?

0%
The states of matter are interconvertible
0%
Force of attraction varies from one kind of matter to another
0%
The states of matter can be changed by changing the temperature of pressure
0%
All of the above

Beer's Law states that

A = a b c

$A = abc$ , where A is absorbance, a is the molar absorptivity constant, b is the path length, and c is the molar concentration.
Which of the following would be an incorrect application of Beer's Law?

0%
The path length is irrelevant, as long as it is inversely proportional to the molar concentration when determining the concentration of an unknown as compared to a standard curve.
0%
The molar concentration is directly related to absorption of light by the sample and is used to develop a standard curve.
0%
The concentration of an unknown sample can be easily determined from a standard curve of known concentrations plotted against absorbance for a fixed path length.
0%
The use of Beer's Law and a spectrophotometer for a colored sample is a better method of determining molar concentration than visual comparison.

Gas A is a colorless gas and Gas B is orange in color. Consider the absorbance data for gases A and B when they were analyzed through spectroscopy.
Which of the following statements best explains the data?

0%
More energy is required to make Gas A molecules vibrate than Gas B.
0%
The ionization energy of Gas B atom is higher than Gas A.
0%
Gas A has more electrons compared to Gas B.
0%
There are lower energy transitions available in Gas B compared to Gas A.

A man writes his biodata with carbon pencil on the plane paper having mass 150 mg. After writing his biodata, he weighs the written paper and find its mass 152 mg. What is the number of carbon atoms present in the paper?

0%
$1.0036 \times 10^{20}$
0%
$5.02 \times 10^{20}$
0%
$1.0036 \times 10^{23}$
0%
$0.502 \times 10^{20}$

Explanation

Mass of C-atoms

= 152 - 150 = 2

$= 152 - 150 = 2$ mg

Molar mass of C-atoms

= 12 g = 12000

$= 12 g = 12000$ mg

12000 mg of carbon contains

6.022 \times 10^{23}

$6.022 \times 10^{23}$ atoms carbon

∴

$\therefore$ 2 mg of carbon contains

= \frac{6.022 \times 10^{23}}{12000} \times 2

$= \dfrac{6.022 \times 10^{23}}{12000} \times 2$

= 0.0010036 \times 10^{23}

$= 0.0010036 \times 10^{23}$

= 1.0036 \times 10^{20}

$= 1.0036 \times 10^{20}$ atoms

Therefore, option A is correct.

The predominant yellow line in the spectrum of a sodium vapour lamp has a wavelength of

590 n m

$590 nm$ . What minimum accelerated potential is needed to excite this line in an electron tube having sodium vapours?

0%
$8.44$ volt.
0%
$2.11$ volt.
0%
$6.11$ volt.
0%
$4.22$ volt.

Explanation

Energy

= \frac{h c}{λ} = \frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{590 \times 10^{- 9}} = 3.369 \times 10^{- 19} J

$=\cfrac{hc}{\lambda}=\cfrac{6.626\times 10^{-34}\times 3\times 10^8}{590\times 10^{-9}}=3.369\times 10^{-19}J$

Ex-potential

=

$=$ Energy.

∴

$\therefore$ Potential required (min)

= \frac{3.369 \times 10^{- 19}}{1.6 \times 10^{- 19}} v \approx 2.11 v

$=\cfrac{3.369\times 10^{-19}}{1.6\times 10^{-19}}v\approx 2.11v$

Which of the following orbitals has the highest energy?

0%
$5d$
0%
$5f$
0%
$6s$
0%
$6p$

Explanation

The orbital energies are compared with the help of

n + l

$n+l$ rule. The higher value of

n + l

$n+l$ , higher is the energy of the orbital.

5 d = n + l = 7

$5d = n+l =7$

5 f = n + l = 8

$5f = n+l =8$

6 s = n + l = 6

$6s = n+l =6$

6 p = n + l = 7

$6p = n+l =7$

The

n + l

$n+l$ value is greatest for 5f orbital.

Hence it has the highest energy among the given options.

In ground state of chromium atom

(Z = 24)

$(Z=24)$ the total number of orbitals populated by one or more electrons is?

0%
$15$
0%
$16$
0%
$20$
0%
$14$

Explanation

The configuration of

_{24} C r

$_{24}Cr$ is

1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{6}, 3 d^{5}, 4 s^{1}

$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 3d^5, 4s^1$

∴

$\therefore$ total s-orbitals

= 4

$=4$
total p-orbitals

= 6

$=6$
total d-orbitals

= 5

$=5$ and thus
Thus, total orbitals

= 4 + 6 + 5 = 15

$=4+6+5=15$ .

The highest energy in Balmer series, in the emission spectra of hydrogen is represented by:

(R_{H} = 109737 {c m}^{- 1})

$\left({R}_{H}=109737 {cm}^{-1}\right)$

0%
$4389.48 {cm}^{-1}$
0%
$2194.74 {cm}^{-1}$
0%
$5486.85 {cm}^{-1}$
0%
$27434.25 {cm}^{-1}$

Explanation

\bar{v} = \frac{1}{λ} = R_{H} Z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]

$\bar{v}=\dfrac{1}{\lambda}=R_H{Z}^{2}\left[\dfrac{1}{{n}_{1}^{2}} - \dfrac{1}{{n}_{2}^{2}}\right]$

= 109737 \times 1^{2} [\frac{1}{2^{2}} - \frac{1}{\infty^{2}}] = 27434.25 {c m}^{- 1}

$=109737 \times {1}^{2} \left[\dfrac{1}{{2}^{2}} - \dfrac{1}{{\infty}^{2}}\right] = 27434.25 {cm}^{-1}$

(series limit will be of highest energy)

The number of revolutions per second made by an electron in the first Bohr's orbit of hydrogen atom is(Given,

r = 0.53 \overset{o}{A}

$r=0.53\overset{o}{A}$ ).

0%
$6\times 10^5$ Hz
0%
$2.5\times 10^{12}$ Hz
0%
$1.9\times 10^{13}$ Hz
0%
$6.5\times 10^{15}$ Hz

Explanation

From Bohr's theory

m v r = \frac{n h}{2 π}

$mvr=\displaystyle\frac{nh}{2\pi}$
For first orbit

n = 1

$n=1$

m (r ω) r = \frac{1 \times h}{2 π}

$m(r\omega )r=\displaystyle\frac{1\times h}{2\pi}$

\Rightarrow m r^{2} 2 π f = \frac{h}{2 π}

$\Rightarrow mr^22\pi f=\displaystyle\frac{h}{2\pi}$

\Rightarrow f = \frac{h}{4 π^{2} m v^{2}}

$\Rightarrow \displaystyle f=\frac{h}{4\pi^2mv^2}$

= \frac{6.6 \times 10^{- 34}}{4 \times (3.14)^{2} \times 9.1 \times 10^{- 31} \times (0.33 \times 10^{- 10})^{2}}

$=\displaystyle\frac{6.6\times 10^{-34}}{4\times (3.14)^2\times 9.1\times 10^{-31}\times (0.33\times 10^{-10})^2}$ .

6.5 \times 10^{15} H z

$6.5\times10^{15} Hz$

The mass of photon having wavelength

1 n m

$1 nm$ is:

0%
$2.21\times { 10 }^{ -35 }kg$
0%
$2.21\times { 10 }^{ -33 }g$
0%
$2.21\times { 10 }^{ -33 }kg$
0%
$2.21\times { 10 }^{ -26 }kg$

Explanation

Photons travel with the speed of light =

3 \times 10^{8} m / s

$3\times 10^8 m/s$

De Broglie wavelength

λ = \frac{h}{m v}

$\lambda = \cfrac{h}{mv}$

h = 6.626 \times 10^{- 34} J s

$h = 6.626 \times 10^{-34} Js$

Now,

1 \times 10^{- 9} = \frac{6.626 \times 10^{- 34}}{m \times 3 \times 10^{8}}

$1 \times 10^{-9} = \cfrac{6.626 \times 10^{-34}}{m \times 3 \times 10^8}$

m = 2.21 \times 10^{- 33} k g

$m = 2.21 \times 10^{-33}kg$

The mass of photon having wavelength 1nm is

2.21 \times 10^{- 33} k g

$2.21 \times 10^{-33}kg$

The fast emission line on hydrogen atomic spectrum in the Balmer series appears at:

[

R =

$R =$ Rydberg constant]

0%
$\dfrac{5R}{36}cm^{-1}$
0%
$\dfrac{3R}{4}cm^{-1}$
0%
$\dfrac{7R}{144}cm^{-1}$
0%
$\dfrac{9R}{400}cm^{-1}$

Explanation

Rydberg - Riz equation is :

\bar{v} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] (W h e r e n_{1} = 2, n_{2} = 3)

$\bar{v}=R_H\left [ \dfrac{1}{n^2_1}-\dfrac{1}{n^2_2} \right ] \,\,\, (Where \, n_1= 2, n_2=3)$

∴ \bar{v} = R_{H} [\frac{9 - 4}{64}] = \frac{5 R}{36} c m^{- 1}

$\therefore \bar{v}=R_H\left [ \dfrac{9-4}{64} \right ] =\dfrac{5R} {36} cm^{-1}$

The magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is:

[

a_{0}

$a_0$ is Bohr radius].

0%
$\dfrac{h^2}{4 \pi^2 ma_o^2}$
0%
$\dfrac{h^2}{16 \pi^2 ma_o^2}$
0%
$\dfrac{h^2}{32 \pi^2 ma_o^2}$
0%
$\dfrac{h^2}{64 \pi^2 ma_o^2}$

Explanation

Solution:- (B)

\frac{h^{2}}{16 π^{2} m a_{o}^{2}}

$\cfrac{{h}^{2}}{16 {\pi}^{2} m a_o^2}$

According to Bohr's model of an atom, the potential energy of electron having charge

- e

$-e$ is given as-

P . E . = \frac{1}{4 π ϵ_{0}} \frac{e}{r} (- e)

$P.E. = \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{e}{r} \left( -e \right)$

\Rightarrow P . E . = - \frac{1}{4 π ϵ_{0}} \frac{e^{2}}{r}

$\Rightarrow P.E. =- \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{{e}^{2}}{r}$

∵ r = \frac{ϵ_{0} n^{2} h^{2}}{π m e^{2}}

$\because r = \cfrac{{\epsilon}_{0} {n}^{2} {h}^{2}}{\pi m {e}^{2}}$

∴ P . E . = \frac{1}{4 π ϵ_{0}} \frac{e^{2}}{(\frac{ϵ_{0} n^{2} h^{2}}{π m e^{2}})}

$\therefore P.E. = \cfrac{1}{4 \pi {\epsilon}_{0}} \cfrac{{e}^{2}}{\left( \cfrac{{\epsilon}_{0} {n}^{2} {h}^{2}}{\pi m {e}^{2}} \right)}$

\Rightarrow P . E . = \frac{1}{4 {ϵ_{0}}^{2}} \frac{m e^{4}}{n^{2} h^{2}}

$\Rightarrow P.E. = \cfrac{1}{4 {{\epsilon}_{0}}^{2}} \cfrac{m {e}^{4}}{{n}^{2} {h}^{2}}$

Now,

a_{0} = \frac{ϵ_{0} h^{2}}{π m e^{2}}

${a}_{0} = \cfrac{{\epsilon}_{0} {h}^{2}}{\pi m {e}^{2}}$

\Rightarrow e^{4} = \frac{{ϵ_{0}}^{2} h^{4}}{π^{2} m^{2} {a_{0}}^{2}}

$\Rightarrow {e}^{4} = \cfrac{{{\epsilon}_{0}}^{2} {h}^{4}}{{\pi}^{2} {m}^{2} {{a}_{0}}^{2}}$

Therefore,

∴ P . E . = \frac{1}{4 {ϵ_{0}}^{2}} \frac{m}{n^{2} h^{2}} (\frac{{ϵ_{0}}^{2} h^{4}}{π^{2} m^{2} {a_{0}}^{2}})

$\therefore P.E. = \cfrac{1}{4 {{\epsilon}_{0}}^{2}} \cfrac{m}{{n}^{2} {h}^{2}} \left( \cfrac{{{\epsilon}_{0}}^{2} {h}^{4}}{{\pi}^{2} {m}^{2} {{a}_{0}}^{2}} \right)$

\Rightarrow P . E . = \frac{h^{2}}{4 π^{2} n^{2} m {a_{0}}^{2}}

$\Rightarrow P.E. = \cfrac{{h}^{2}}{4 {\pi}^{2} {n}^{2} m {{a}_{0}}^{2}}$

For second orbit of hydrogen atom,

n = 2

$n = 2$

\Rightarrow P . E . = \frac{h^{2}}{16 π^{2} m {a_{0}}^{2}}

$\Rightarrow P.E. = \cfrac{{h}^{2}}{16 {\pi}^{2} m {{a}_{0}}^{2}}$

Hence the magnitude of potential energy of an electron in the second Bohr orbit of a hydrogen atom is

\frac{h^{2}}{16 π^{2} m {a_{0}}^{2}}

$\cfrac{{h}^{2}}{16 {\pi}^{2} m {{a}_{0}}^{2}}$ .

The figure shown the wavelength spectrum of x-rays produced when

50 K e V

$50 \ KeV$ electrons strike a molybdenum target. The

K_{α}

${ K }_{ \alpha }$ line on the figure is represent by a number:

0%
$1$
0%
$2$
0%
$3$
0%
none

Explanation

K_{α}

$K_{\alpha}$ refers to the peak due to

e^{-}

$e^{-}$ transition from

n = 1

$n=1$ orbit

Ans is

(A)

$(A)$

Calculate the wavelength of the first line of

H e^{+}

$He^{+}$ ion spectral series whose interval with extreme lines is:

\frac{1}{λ_{1}} - \frac{1}{λ_{2}} = 2.7541 \times 10^{4} c m^{- 1}

$\dfrac{1}{\lambda_{1}} - \dfrac{1}{\lambda_{2}} = 2.7541 \times 10^{4} cm^{-1}$

0%
$4869 \overset{o}{A}$
0%
$1651 \overset{o}{A}$
0%
$3039 \overset{o}{A}$
0%
$2175 \overset{o}{A}$

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

0%
$He^+$
0%
$Li^{2+}$
0%
$Be^{3+}$
0%
All of the these

Explanation

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

H e^{+}, L i^{2 +}, B e^{3 +}

$He^+, Li^{2+},Be^{3+}$ etc.

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

0%
$\frac{4\pi^2 mr^2}{nh}$
0%
$\frac{nh}{4\pi^2 mr}$
0%
$\frac{2\pi mr}{n^2h^2}$
0%
$\frac{h}{2\pi mr}$

Explanation

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

If the radial probability curve indicates

2 s

$2s$ orbital, the distance between the peak points X and Y is?

0%
$2.07\overset{o}{A}$
0%
$1.59\overset{o}{A}$
0%
$0.53\overset{o}{A}$
0%
$2.12\overset{o}{A}$

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Structure Of Atom Quiz 12 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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