Structure of Atom — Page 4 | JEE Chemistry

Q31

Ionisation energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is

A 4.41 x 10–16 J atom–1

B -4.41 x 10–17 J atom–1

C -2.2 x 10–15 J atom–1

D 8.82 x 10–17 J atom–1

Correct Answer

Option B

Solution

Ionisation energy(IE) - It is the energy required to move an electron from ground state to infinity. IE =

{E_\infty } - {E_1}

=

0 - {E_1}

=

- {E_1}

$\therefore$ E1 of He+ = - 19.6 x 10–18 J atom–1 Energy of a species at n state, (En)species = (En)hydrogen $\times$ Z2 $\therefore$ (E1)hydrogen =

{{ - 19.6 \times {{10}^{ - 18}}} \over 4}

[ For He, Z = 2 ] (E1)Li+2 =

{{ - 19.6 \times {{10}^{ - 18}}} \over 4} \times {3^2}

= -4.41 x 10–17 J atom–1

Q32

The total number of orbitals associated with the principal quantum number 5 is :

A 5

B 10

C 20

D 25

Correct Answer

Option D

Solution

Total number of orbitals associated with principal quantum number n is = n2.

Here, n = 5.

$\therefore$ The total number of orbitals associated with the principal quantum number 5 is = 52 = 25

Q33

The isoelectronic set of ions is :

A F–, Li+, Na+ and Mg2+

B Li+, Na+, O2- and F-

C N3–, O2-, F- and Na+

D N3–, Li+, Mg2+ and O2-

Correct Answer

Option C

Solution

Atomic numbers of N, O, F and Na are 7, 8, 9 and 11 respectively.

Therefore, total number of electrons in each of N3-, O2–, F– and Na+ is 10 and hence they are isoelectronic.

Q34

The radius of the second Bohr orbit for hydrogen atom is: (Planck’s Const. h = 6.6262 × 10-34 Js; mass of electron = 9.1091 × 10-31 kg; charge of electron (e) = 1.60210 × 10-19 C; permittivity of vacuum (

{\varepsilon _0}

) = 8.854185 × 10-12 kg-1 m-3 A2)

A 4.76

\mathop {\rm A}\limits^o

B 2.12

\mathop {\rm A}\limits^o

C 0.529

\mathop {\rm A}\limits^o

D 1.65

\mathop {\rm A}\limits^o

Correct Answer

Option B

Solution

Radius of an atom in nth orbit, rn = 0.529

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

Here n = 2 and for hydrogen atom, atomic number (Z) = 1 $\therefore$ r2 = 0.529

\times {{{2^2}} \over 1}

= 2.12

\mathop {\rm A}\limits^o

Q35

The upper stratosphere consisting of the ozone layer protects us from sun's radiation that falls in the wavelength region of -

A 200-315 nm

B 400-550 nm

C 0.8-1.5 nm

D 600-750 nm

Correct Answer

Option A

Solution

Sun emits UV-radiations, which have the wavelength range from 1 nm to 400 nm.

Q36

If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength

\lambda

, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

A 1/2

\lambda

B 3/4

\lambda

C 4/9

\lambda

D 2/3

\lambda

Correct Answer

Option C

Solution

From photoelectric effect, E = $\phi$ + KE

{{hc} \over \lambda } = \phi + {{{p^2}} \over {2m}}

................ (1) Now when momentum = 1.5p then let wavelength =

{{\lambda _1}}

$\therefore$

{{hc} \over {{\lambda _1}}} = \phi + {{{{\left( {1.5p} \right)}^2}} \over {2m}}

...........

(2) Given, kinetic energy(KE) of ejected photoelectron to be very high in comparison to work function( $\phi$ ).

$\therefore$ We can neglect work function( $\phi$ ).

$\therefore$ Equation (1) and (2) becomes,

{{hc} \over \lambda } = {{{p^2}} \over {2m}}

................ (1)

{{hc} \over {{\lambda _1}}} = {{{{\left( {1.5p} \right)}^2}} \over {2m}}

........... (2) Dividing (1) by (2) we get,

{{{\lambda _1}} \over \lambda } = {{{p^2}} \over {{{\left( {1.5p} \right)}^2}}}

$\Rightarrow$

{{{\lambda _1}} \over \lambda } = {4 \over 9}

$\Rightarrow$

{\lambda _1} = {4 \over 9}\lambda

Q37

Which of the following combination of statements is true regarding the interpretation of the atomic orbitals ? (a) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. (b) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number. (c) According to wave mechanics, the ground state angular momentum is equal to

{h \over {2\pi }}

. (d) The plot of

\psi

vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value.

A (a), (b)

B (a), (d)

C (b), (c)

D (a), (c)

Correct Answer

Option B

Solution

(a) Angular momentum,

mvr = {{nh} \over {2\pi }}

\Rightarrow mvr \propto n

$\propto$ distance from the nucleus (b) This statement is incorrect as size of an orbit $\propto$ Azimuthal quantum number (l) ( $\because$ n = constant) (c) This statement is incorrect as at ground state, n = 1, l = 0 $\Rightarrow$ Orbital angular momentum (wave mechanics)

= \sqrt {l(l + 1)} {h \over {2\pi }} = 0

[ $\because$

l = 0

] (d) The plot of

\psi

vs

r

for various azimuthal quantum numbers, shows peak shifting towards higher

r

value. This statement is typically true. The radial distribution functions of atomic orbitals (which can be related to

|\psi|^2

vs

r

) for higher azimuthal quantum numbers generally have their peaks at larger values of the radial distance

r

from the nucleus. This is because with higher

l

values, the electrons are more likely to be found at greater distances due to the higher angular momentum.The given plot is

Q38

The ground state energy of hydrogen atom is – 13.6 eV. The energy of second excited state of He+ ion in eV is :

A

-

6.04

B

-

54.4

C

-

27.2

D

-

3.4

Correct Answer

Option A

Solution

(E)nth = (EGND)H .

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

E3rd (He+) = ( $-$ 13.6eV) .

{{{2^2}} \over {{3^2}}}

= $-$ 6.04 eV

Q39

Which of the following forms of hydrogen emits low energy

\beta

- particles?

A Tritium

_1^3

H

B Proton H+

C Protium

_1^1

H

D Deuterium

_1^2

H

Correct Answer

Option A

Solution

Tritium isotope of hydrogen is radioactive and emits low energy

\beta ^-

particles. It is because of high n/p ratio of tritium which makes nucleus unstable.

Q40

A certain orbital has no angular nodes and two radial nodes. The orbital is :

A 2s

B 3s

C 3p

D 2p

Correct Answer

Option B

Solution

Angular nodes =

\ell

= 0 Radial nodes $\Rightarrow$ (n –

\ell

– 1) = 2 $\Rightarrow$ n – 0 – 1 = 2 $\Rightarrow$ n = 3 So orbital is 3s.