Structure of Atom — Page 3 | JEE Chemistry

Q21

Given below are two statements: Statement I : According to Bohr's model of hydrogen atom, the angular momentum of an electron in a given stationary state is quantised. Statement II : The concept of electron in Bohr's orbit, violates the Heisenberg uncertainty principle. In the light of the above statements, choose the most appropriate answer from the options given below:

A Both Statement I and Statement II are incorrect

B Statement I is correct but Statement II is incorrect

C Both Statement I and Statement II are correct

D Statement I is incorrect but Statement II is correct

Correct Answer

Option C

Solution

Both Statement I and Statement II are correct Statement I is correct.

According to Bohr's model of the hydrogen atom, the angular momentum of an electron in a given stationary state is quantized.

The angular momentum of an electron in the nth orbit is given by: $L = n\hbar = n \dfrac{h}{2\pi}$ where n is an integer (quantum number), and h is the Planck's constant.

Statement II is also correct.

Bohr's model does not consider the Heisenberg uncertainty principle, which states that it is impossible to know the exact position and momentum of an electron simultaneously.

In Bohr's model, electrons are assumed to move in well-defined orbits with quantized angular momentum, which implies knowledge of both position and momentum, thus violating the Heisenberg uncertainty principle.

Q22

The shortest wavelength of H atom in the Lyman series is

\lambda

1. The longest wavelength in the Balmar series of He+ is :

A

{{5{\lambda _1}} \over 9}

B

{{36{\lambda _1}} \over 5}

C

{{27{\lambda _1}} \over 5}

D

{{9{\lambda _1}} \over 5}

Correct Answer

Option D

Solution

For Shortest wavelength energy should be maximum. For maximum energy transition must be form n = $\infty$ to n = 1.

{1 \over {{\lambda _1}}} = {R_H}{\left( 1 \right)^2}\left[ {{1 \over 1} - 0} \right]

$\Rightarrow$

{1 \over {{\lambda _1}}} = {R_H}

For longest wavelength,

\Delta

E = minimum. For Blamer series n = 3 to n = 2 will have

\Delta

E minimum for He+, Z = 2 So

{1 \over {{\lambda _2}}} = {R_H}{\left( 2 \right)^2}\left[ {{1 \over 4} - {1 \over 9}} \right]

$\Rightarrow$

{1 \over {{\lambda _2}}} = {R_H} \times {5 \over 9}

$\Rightarrow$

{\lambda _2} = {\lambda _1} \times {9 \over 5}

Q23

If the radius of the 3rd Bohr's orbit of hydrogen atom is r3 and the radius of 4th Bohr's orbit is r4. Then :

A

{r_4} = {9 \over {16}}{r_3}

B

{r_4} = {16 \over {9}}{r_3}

C

{r_4} = {3 \over {4}}{r_3}

D

{r_4} = {4 \over {3}}{r_3}

Correct Answer

Option B

Solution

We know,

r = {r_0} \times {{{n^2}} \over z}

For hydrogen atom, $\therefore$

{r_3} = {r_0} \times {{{3^2}} \over 1}

\Rightarrow {r_0} \times {{{r_3}} \over 9}

and

{r_4} = {r_0} \times {{{4^2}} \over 1}

= {{{r_3}} \over 9} \times 16

= {{16} \over 9}{r_3}

Q24

According to Bohr's model of hydrogen atom, which of the following statement is incorrect?

A Radius of

4^{\text{th }}

orbit is four times larger than that of

2^{\text{nd }}

orbit

B Radius of

8^{\text{th }}

orbit is four times larger than that of

4^{\text{th }}

orbit

C Radius of

6^{\text{th }}

orbit is three times larger than that of

4^{\text{th }}

orbit

D Radius of

3^{\text{rd }}

orbit is nine times larger than that of

1^{\text{st }}

orbit

Correct Answer

Option C

Solution

In Bohr’s model the radius of the $n$ th orbit is

r_n = n^2\,a_0

where $a_0$ is the Bohr radius. Hence for any two orbits $m$ and $n$ :

\frac{r_m}{r_n} = \frac{m^2}{n^2}\,.

Let’s check each option: A: $\displaystyle \dfrac{r_4}{r_2} = \dfrac{4^2}{2^2} = \dfrac{16}{4} = 4$ → correct B: $\displaystyle \dfrac{r_8}{r_4} = \dfrac{8^2}{4^2} = \dfrac{64}{16} = 4$ → correct C: $\displaystyle \dfrac{r_6}{r_4} = \dfrac{6^2}{4^2} = \dfrac{36}{16} = 2.25\;\bigl(\neq3\bigr)$ → incorrect D: $\displaystyle \dfrac{r_3}{r_1} = \dfrac{3^2}{1^2} = 9$ → correct Therefore the incorrect statement is Option C.

Q25

The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency '

A

'

\times 10^{12}

hertz and that has a radiant intensity in that direction of

\dfrac{1}{{ 'B'}}

, watt per steradian. '

A

' and '

B

' are respectively

A 450 and 683

B 540 and

\dfrac{1}{683}

C 540 and 683

D 450 and

\dfrac{1}{683}

Correct Answer

Option C

Solution

The definition of the candela involves a source emitting monochromatic radiation of a specific frequency and having a precise radiant intensity in a given direction.

The candela is thereby defined based on both the frequency of the monochromatic light and its radiant intensity in watts per steradian.

According to the given statement, we are dealing with a frequency identified as '

A

'

\times 10^{12}

hertz and a radiant intensity specified as

\frac{1}{B}

watt per steradian.

However, there seems to be a small confusion in how the values are represented in the question.

The correct interpretation and formulation based on the International System of Units (SI) are as follows: The frequency of the monochromatic radiation that is used in defining the candela is 540 THz (tera hertz), so the value of

A

should be 540.

The specific radiant intensity mentioned relates to how the candela is defined in terms of a source's power output in a particular direction within a given frequency.

The value that corresponds to the efficiency of human visual perception at this frequency is approximately 683 lumens per watt.

Therefore, the number 683 is involved in specifying how many lumens correspond to a power of one watt of monochromatic light at 540 THz.

Given this context, the correct answer derives from understanding that: '

A

' represents the frequency in THz, which is 540 for the definition of the candela. '

B

' refers to the conversion factor or the luminous efficacy in lumens per watt at this frequency, which is 683 lumens per watt, not its reciprocal.

Therefore, the correct values for '

A

' and '

B

' are 540 and 683, respectively, making Option C the correct answer.

Q26

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Energy of

2 \mathrm{s}

orbital of hydrogen atom is greater than that of

2 \mathrm{s}

orbital of lithium. Reason R: Energies of the orbitals in the same subshell decrease with increase in the atomic number. In the light of the above statements, choose the correct answer from the options given below.

A Both A and R are true and R is the correct explanation of A.

B Both A and R are true but R is NOT the correct explanation of A.

C A is true but R is false.

D A is false but R is true.

Correct Answer

Option A

Solution

As the atomic number increases then the potential energy of electrons present in same shell becomes more and more negative.

And therefore total energy also becomes more negative.

\mathrm{E}_{\text{total }}=-13.6 \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}

$\therefore$ Energies of the orbitals in the same subshell decreases with increase in atomic number.

Q27

According to Bohr's atomic theory : (A) Kinetic energy of electron is

\propto {{{Z^2}} \over {{n^2}}}

. (B) The product of velocity (v) of electron and principal quantum number (n),

'vn' \propto {Z^2}

. (C) Frequency of revolution of electron in an orbit is

\propto {{{Z^3}} \over {{n^3}}}

. (D) Coulombic force of attraction on the electron is

\propto {{{Z^3}} \over {{n^4}}}

. Choose the most appropriate answer from the options given below :

A (A), (C) and (D) only

B (A) only

C (C) only

D (A) and (D) only

Correct Answer

Option D

Solution

According to Bohr's theory, I.

KE \propto {{{Z^2}} \over {{n^2}}}

or

13.6 \propto {{{Z^2}} \over {{n^2}}}{{(eV)} \over {(atom)'}}

( $\therefore$ Correct) II. Speed of electron

\propto {Z \over n}

(Here, Z = atomic number, n = number of shells) $\therefore$

v \times n \propto Z

( $\therefore$ Incorrect) III. Frequency of revolution of electron

= {v \over {2\pi r}}

Frequency

\propto {{{Z^2}} \over {{n^3}}}

( $\because$

v \propto {z \over n},r \propto {{{n^2}} \over z}

) ( $\therefore$ Incorrect)

Q28

The number of radial node/s for

3 p

orbital is :

A 3

B 2

C 1

D 4

Correct Answer

Option C

Solution

The number of radial nodes in an orbital is given by the formula:

\text{Number of radial nodes} = n - l - 1

where

n

is the principal quantum number,

l

is the azimuthal quantum number also known as the angular momentum quantum number. For a

3p

orbital,

n = 3

(since it is the 3rd energy level),

l = 1

(since

p

orbitals correspond to

l = 1

). Let's use the formula to determine the number of radial nodes:

\text{Number of radial nodes} = 3 - 1 - 1 = 1

So, the correct answer is: Option C : 1

Q29

If the radius of the first orbit of hydrogen atom is

\alpha_{0}

, then de Broglie's wavelength of electron in

3^{\text{rd }}

orbit is :

A

\dfrac{\pi \alpha^{3}}{6}

B

3\pi\alpha_0

C

6\pi\alpha_0

D

\dfrac{\pi \alpha^{3}}{3}

Correct Answer

Option C

Solution

In Bohr's model, the angular momentum of an electron in an orbit is quantized, which can be represented as:

mvr = n \cdot \frac{h}{2\pi}

Where: (m) is the mass of the electron, (v) is the velocity of the electron, (r) is the radius of the nth orbit, and (n) is the principal quantum number ((n = 1) for the first orbit, (n = 2) for the second orbit, and so on).

The de Broglie wavelength ( $\lambda$ ) is given by:

\lambda = \frac{h}{mv}

For the nth orbit, the radius (r) is $n^2$ times the radius of the first orbit ( $r_0$ , or $\alpha_0$ in your notation):

r = n^2 \cdot \alpha_0

The circumference of the nth orbit is $2\pi r$ , which can be written as:

2\pi r = n \cdot \lambda

Substituting the expression for (r):

2\pi \cdot n^2 \cdot \alpha_0 = n \cdot \lambda

Solving for $\lambda$ gives:

\lambda = 2\pi \cdot n \cdot \alpha_0

For the 3rd orbit ((n = 3)), this simplifies to:

\lambda = 6\pi \cdot \alpha_0

Q30

Uncertainty in position of a minute particle of mass 25 g in space is 10-5 m. What is the uncertainty in its velocity (in ms-1) (h = 6.6

\times

10-34 Js)

A 2.4

\times

10-34

B 0.5

\times

10-34

C 2.1

\times

10-28

D 0.5

\times

10-23

Correct Answer

Option C

Solution

According to Hysenberg's Uncertainty Principal

\Delta x . \Delta p \ge { h \over {4\pi}}

$ Where

\Delta x

is uncertainty in position and

\Delta p

is uncertainty in momentum and as in momentum m (mass of object) is always same so we can say

\Delta x . m. \Delta v \ge { h \over {4\pi}}

$ Mass of the particle (m) = 25 g = 0.025 kg Position of the particle (

\Delta x

) = 10-5 m

\Delta x . \Delta p

=

{ h \over {4\pi}}

\Delta x . m . \Delta v

=

{ h \over {4\pi}}

\Delta v

=

{h \over {4 \times \Delta x \times m \times \pi }}

\Delta v

=

{{6.6 \times {{10}^{ - 34}}} \over {4 \times {{10}^{ - 5}} \times 0.025 \times 3.14}}

\Delta v

= 2.1 $\times$ 10-28