Structure of Atom — Page 9 | JEE Chemistry

Q81

Which of the following is the energy of a possible excited state of hydrogen?

A –6.8 eV

B –3.4 eV

C +6.8 eV

D +13.6 eV

Correct Answer

Option B

Solution

Energy(En) =

- {{13.6{Z^2}} \over {{n^2}}}eV

where n = 1, 2, 3, 4, ...... For hydrogen Z = 1 and excited state starts from n = 2 So putting n = 2 E2 =

- {{13.6} \over {{2^2}}}eV

= -3.4 eV

Q82

Hydrogen has three isotopes (A), (B) and (C). If the number of neutron(s) in (A), (B) and (C) respectively, are (x), (y) and (z), the sum of (x), (y) an (z) is :

A 3

B 1

C 4

D 2

Correct Answer

Option A

Solution

Hydrogen has three isotopes (A) Protium (

{}_1^1H

) has 0 neutron. (B) Deutrium (

{}_1^2H

) has 1 neutrons. (C) Tritium (

{}_1^3H

) has 2 neutrons. Total number of neutrons in three isotopes of hydrogen = 0 + 1 + 2 = 3

Q83

The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with :

A Lyman series

B Balmer series

C Paschen series

D Brackett series

Correct Answer

Option B

Solution

Here electron is coming to the orbital of radius 211.6 pm.

Now we have to find which series have radius of orbital 211.6 pm.

We know, Radius, r =

0.529 \times {{{n^2}} \over Z}\,\mathop A\limits^ \circ

Given, r = 211.6 pm = 211.6 $\times$ 10-12 m and Z = 1 for hydrogen atom $\therefore$ 211.6 $\times$ 10-12 =

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

$\times$ 10-10 $\Rightarrow$ n = 2 As n = 2 so the series is Balmer Series.

Q84

The de Broglie wavelength (

\lambda

) associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency] :

A

\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{3 \over 2}}}}}

B

\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 4}}}}}

C

\lambda \,\infty \,{1 \over {\left( {v - {v_0}} \right)}}

D

\lambda \,\infty \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}

Correct Answer

Option D

Solution

By photoelectric effect KE = h $\gamma$ - h $\gamma$ o ....(1) de broglie wavelength, $\lambda$ =

{h \over {mv}}

=

{h \over {\sqrt {2m \times K.E} }}

...(2) Using equation (1) and (2), we get $\lambda$ =

{h \over {\sqrt {2m \times \left( {h\nu - h{\nu _0}} \right)} }}

$\therefore$

\lambda \,\propto \,{1 \over {{{\left( {v - {v_0}} \right)}^{{1 \over 2}}}}}

Q85

For the Balmer series in the spectrum of H atom,

\overline \nu = {R_H}\left\{ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right\}

, the correct statements among (I) to (IV) are : (I) As wavelength decreases, the lines in the series converge (II) The integer n1 is equal to 2 (III) The lines of longest wavelength corresponds to n2 = 3 (IV) The ionization energy of hydrogen can be calculated from wave number of these lines

A (II), (III), (IV)

B (I), (II), (III)

C (I), (III), (IV)

D (I), (II), (IV)

Correct Answer

Option B

Solution

For balmer series : n1 = 2, n2 = 3, 4, 5, ..... $\infty$ For longest wavelength n2 = 3

{1 \over \lambda } = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)

As wavelength decreases the lines in the Balmer series converge. The correct statements are (I), (II) and (III).

Q86

The radius of the second Bohr orbit, in terms of the Bohr radius, a0, in Li2+ is :

A

{{2{a_0}} \over 9}

B

{{2{a_0}} \over 3}

C

{{4{a_0}} \over 9}

D

{{4{a_0}} \over 3}

Correct Answer

Option D

Solution

{r_n} = {{{n^2}{a_0}} \over Z}

For 2nd Bohr orbit of Li+2 n = 2 and Z = 3 r =

{{{2^2}{a_0}} \over 3}

=

{{4{a_0}} \over 3}

Q87

Which of the following sets of quantum numbers is not allowed?

A

\mathrm{n}=3,1=2, \mathrm{~m}_{\mathrm{l}}=0, \mathrm{~s}=+\dfrac{1}{2}

B

\mathrm{n}=3, \mathrm{l}=2, \mathrm{~m}_{\mathrm{l}}=-2, \mathrm{~s}=+\dfrac{1}{2}

C

\mathrm{n}=3, \mathrm{l}=3, \mathrm{~m}_{\mathrm{l}}=-3, \mathrm{~s}=-\dfrac{1}{2}

D

\mathrm{n}=3, \mathrm{l}=0, \mathrm{~m}_{1}=0, \mathrm{~s}=-\dfrac{1}{2}

Correct Answer

Option C

Solution

If

n=3

, then possible values of

l=0,1,2

But in option (C), the value of

\mathrm{l}

is given ' 3 ', this is not possible.

Q88

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from

\mathrm{n=4}

to

\mathrm{n}=2

of

\mathrm{He}^{+}

spectrum

A

\mathrm{n}=3

to

\mathrm{n}=4

B

\mathrm{n}=2

to

\mathrm{n}=1

C

\mathrm{n}=1

to

\mathrm{n}=2

D

\mathrm{n}=1

to

\mathrm{n}=3

Correct Answer

Option B

Solution

$\mathrm{He}^{+}$ ion :

\begin{aligned} & \frac{1}{\lambda(\mathrm{H})}=\mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\\\ & \frac{1}{\lambda\left(\mathrm{He}^{+}\right)}=\mathrm{R}(2)^2\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \text{ Given } \lambda(\mathrm{H})=\lambda\left(\mathrm{He}^{+}\right) \\\\ & \mathrm{R}(1)^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right]=\mathrm{R}(4)\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\\\ & \frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}=\frac{1}{1^2}-\frac{1}{2^2} \end{aligned}

On comparing $\mathrm{n}_1=1$ and $\mathrm{n}_2=2$ .

Q89

Given below are two statements : Statement (I) : For a given shell, the total number of allowed orbitals is given by

n^2

. Statement (II) : For any subshell, the spatial orientation of the orbitals is given by

-l

to

+l

values including zero. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option A

Solution

For a shell total number of orbitals $=\mathrm{n}^2$ Magnetic quantum number have values $(-\ell$ to $+\ell$ ) including 0.

Q90

The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is :

A 0.529

\mathop {\rm A}\limits^ \circ

B 2

\pi

\times

0.529

\mathop {\rm A}\limits^ \circ

C

{{0.529} \over {2\pi }}

\mathop {\rm A}\limits^ \circ

D 4

\times

0.529

\mathop {\rm A}\limits^ \circ

Correct Answer

Option B

Solution

Radius r = 0.529 $\times$

{{{n^2}} \over z}

In first Bohr orbit of hydrogen atom radius, r = 0.529 $\times$

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

= 0.529

\mathop A\limits^ \circ

Angular momentum, mvr =

{{nh} \over {2\pi }}

\therefore\,\,\,

for n = 1, mvr =

{h \over {2\pi }}

$\Rightarrow$

\,\,\,

2 $\pi$ r =

{h \over {mv}}

= $\lambda$ [ as $\lambda$ =

{h \over {mv}}

]

\therefore\,\,\,

$\lambda$ = 2 $\pi$ r = 2 $\pi$ $\times$ 0.5 29