Structure of Atom — Page 5 | NEET Chemistry

Q41

If uncertainty in position and momentum are equal, then uncertainty in velocity is

A

{1 \over m}\sqrt {{h \over \pi }}

B

\sqrt {{h \over \pi }}

C

{1 \over {2m}}\sqrt {{h \over \pi }}

D

\sqrt {{h \over {2\pi }}}

Correct Answer

Option C

Solution

According to Heisenberg uncertainty principle

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

or m

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

$\Rightarrow$ (m

\Delta v

) 2

\ge {h \over {4\pi }}

( $\because$

\Delta x

=

\Delta p

) $\Rightarrow$

\Delta v \ge

{1 \over {2m}}\sqrt {{h \over \pi }}

Q42

Consider the following sets of quantum numbers : <br><br> <table class=tg> <tbody><tr> <th class=tg-wv9z></th> <th class=tg-wv9z>n</th> <th class=tg-baqh></th> <th class=tg-wv9z>l</th> <th class=tg-baqh></th> <th class=tg-wv9z>m</th> <th class=tg-baqh></th> <th class=tg-wv9z>s</th> </tr> <tr> <td class=tg-wv9z>(i)</td> <td class=tg-wv9z>3</td> <td class=tg-baqh></td> <td class=tg-wv9z>0</td> <td class=tg-baqh></td> <td class=tg-wv9z>0</td> <td class=tg-baqh></td> <td class=tg-wv9z>+1/2</td> </tr> <tr> <td class=tg-wv9z>(ii)</td> <td class=tg-wv9z>2</td> <td class=tg-baqh></td> <td class=tg-wv9z>2</td> <td class=tg-baqh></td> <td class=tg-wv9z>1</td> <td class=tg-baqh></td> <td class=tg-wv9z>+1/2</td> </tr> <tr> <td class=tg-wv9z>(iii)</td> <td class=tg-wv9z>4</td> <td class=tg-baqh></td> <td class=tg-wv9z>3</td> <td class=tg-baqh></td> <td class=tg-wv9z>-2</td> <td class=tg-baqh></td> <td class=tg-wv9z>-1/2</td> </tr> <tr> <td class=tg-wv9z>(iv)</td> <td class=tg-wv9z>1</td> <td class=tg-baqh></td> <td class=tg-wv9z>0</td> <td class=tg-baqh></td> <td class=tg-wv9z>-1</td> <td class=tg-baqh></td> <td class=tg-wv9z>-1/2</td> </tr> <tr> <td class=tg-wrg0>(v)</td> <td class=tg-wrg0>3</td> <td class=tg-baqh></td> <td class=tg-wrg0>2</td> <td class=tg-baqh></td> <td class=tg-wrg0>3</td> <td class=tg-baqh></td> <td class=tg-wrg0>+1/2</td> </tr> </tbody></table> <br><br>Which of the following sets of quantum number is not possible ?

A (i), (ii), (iii) and (iv)

B (ii), (iv) and (v)

C (i) and (iii)

D (ii), (iii) and (iv)

Correct Answer

Option B

Solution

(i) represents an electron in 3s orbital (ii) is not possible as value of m varies from 0, 1, .... (

n

-1) (iii) represents an electron in 4f orbital (iv) is not possible as value of m varies from -

l

... +

l

(v) is not possible as value of m varies from -

l

... +

l

, it can never be grater than

l

Q43

Given : The mass of electron is 9.11

\times

10

-

31 kg, Planck constant is 6.626

\times

10

-

34 J s, the uncertainty involved in the measurement of velocity within a distance of 0.1

\mathop A\limits^ \circ

is

A 5.79

\times

10 5 m s

-

1

B 5.79

\times

10 6 m s

-

1

C 5.79

\times

10 7 m s

-

1

D 5.79

\times

10 8 m s

-

1

Correct Answer

Option B

Solution

\Delta x.m\Delta v = h/4\pi

0.1 $\times$ 10 -10 $\times$ 9.11 $\times$ 10 -31 $\times$

\Delta v

=

{{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143}}

$\therefore$

\Delta v

=

{{6.626 \times {{10}^{ - 34}}} \over {4 \times 3.143 \times 0.1 \times {{10}^{ - 10}} \times 9.11 \times {{10}^{ - 31}}}}

= 5.79 $\times$ 10 6 m s -1

Q44

The orientation of an atomic orbital is governed by

A principal quantum number

B azimuthal quantum number

C spin quantum number

D magnetic quantum number.

Correct Answer

Option D

Solution

Magnetic quantum number describes the orientation.

Q45

The energy of second Bohr orbit of the hydrogen atom is

-

328 kJ mol

-

1 ; hence the energy of fourth Bohr orbit would be

A

-

41 kJ mol

-

1

B

-

82 kJ mol

-

1

C

-

164 kJ mol

-

1

D

-

1312 kJ mol

-

1

Correct Answer

Option B

Solution

E n = -K

{\left( {{Z \over n}} \right)^2}

Z = 1; n = 2 E 2 =

{ {{-K \times 1 \over 4}}}

$\Rightarrow$ E 2 = -328 kJ mol -1 ; K = 4 $\times$ 328 E 4 =

{ {{-K \times 1 \over 16}}}

$\Rightarrow$ E 4 = -4 $\times$ 328

{ {{ 1 \over 16}}}

= -82 kJ mol -1

Q46

The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in hydrogen atom will be (Given ionization energy of H = 2.18

\times

10

-

18 J atom

-

1 and h = 6.625

\times

10

-

34 J s)

A 1.54

\times

10 15 s

-

1

B 1.03

\times

10 15 s

-

1

C 3.08

\times

10 15 s

-

1

D 2.00

\times

10 15 s

-

1

Correct Answer

Option C

Solution

E = h

v

or

v

= E/h For H atom, E =

{{ - 21.76 \times {{10}^{ - 19}}} \over {{n^2}}}J\,at{m^{ - 1}}

\Delta E = - 21.76 \times {10^{ - 19}}\left( {{1 \over {{4^2}}} - {1 \over {{1^2}}}} \right)

= 20.40 $\times$ 10 -19 J atm -1

v

=

{{20.40 \times {{10}^{ - 19}}} \over {6.626 \times {{10}^{ - 34}}}}

= 3.079 $\times$ 10 15 s -1

Q47

The velocity of Planck's constant is 6.63

\times

10

-

34 J s. The velocity of light is 3.0

\times

10 8 m s

-

1 . Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8

\times

10 15 s

-

1 ?

A 2

\times

10

-

25

B 5

\times

10

-

18

C

4 \times {10^1}

D 3

\times

10 7

Correct Answer

Option C

Solution

v

= c/ $\lambda$ $\lambda$ =

{c \over v}

=

{{3 \times {{10}^8}} \over {8 \times {{10}^{15}}}}

= 37.5 $\times$ 10 -9 m = 37.5 nm $\approx$

4 \times {10^1}

nm

Q48

In hydrogen atom, energy of first excited state is

-

3.4 eV. Then find out K.E. of same orbit of hydrogen atom

A +3.4 eV

B +6.8 eV

C

-

13.6 eV

D +13.6 eV

Correct Answer

Option A

Solution

K.E = 1/2 mv 2 =

{\left( {{{\pi {e^2}} \over {nh}}} \right)^2} \times 2m

[ $\because$ v =

{{2\pi {e^2}} \over {nh}}

] $\therefore$ Total energy = E n =

{{ - 2{\pi ^2}m{e^4}} \over {{n^2}{h^2}}}

= -

{\left( {{{\pi {e^2}} \over {nh}}} \right)^2}

$\times$ 2m = -K.E $\therefore$ K.E = - E n Energy of first excited state is -3.4 eV $\therefore$ Kinetic energy of the same orbit (n = 2) will be +3.4 ev

Q49

The following quantum numbers are possible for how many orbitals : n = 3,

l

= 2, m = +2 ?

A 1

B 2

C 3

D 4

Correct Answer

Option A

Solution

n = 3, l = 2, m = +2 It shows one of the five d-orbitals(3d)

Q50

Main axis of a diatomic molecule is z, molecular orbital p x and p y overlap to from which of the following orbitals.

A

\pi

molecular orbital

B

\sigma

molecular orbital

C

\delta

molecular orbital

D No bond will form

Correct Answer

Option A

Solution

For $\pi$ overlap, the lobes of the atomic orbitals are perpendicularto the line joining the nuclei.

Hence only sidewise overlapping takes place.