Atoms and Nuclei Questions — JEE Physics

Q1

Which of the following cannot be emitted by radioactive substances during their decay ?

A Protons

B Neutrinoes

C Helium nuclei

D Electrons

Correct Answer

Option A

Solution

The radioactive substances emit $\alpha$ -particles (Helium nucleus), $\beta$ -particles (electrons) and neutrinoes.

Q2

The ratio of mass densities of nuclei of 40Ca and 16O is close to :-

A 1

B 5

C 0.1

D 2

Correct Answer

Option A

Solution

Densities of nucleus happens to be constant, irrespective of mass number.

Q3

Which of the following atoms has the lowest ionization potential ?

A

{}_7^{14}N

B

{}_{55}^{133}\,Cs

C

{}_{18}^{40}\,Ar

D

{}_8^{16}\,O

Correct Answer

Option B

Solution

The ionisation potential increases from left to right in a period and decreases from top to bottom in a group.

Therefore ceasium will have the lowest ionisation potential.

Q4

A nuclear transformation is denoted by

X\left( {n,\alpha } \right)\begin{array}{ll}7 \\ 3 \end{array} Li.

Which of the following is the nucleus of element

X

?

A

{}_5^{10}B

B

{}^{12}{C_6}

C

{}_4^{11}Be

D

{}_5^9B

Correct Answer

Option A

Solution

{}_z{X^A} + {}_0{n^1} \to {}_3L{i^7} + {}_2H{e^4}

On comparison,

A = 7 + 4 - 1 = 10,\,\,\,z = 3 + 2 - 0 = 5

It is boron

{}_5{B^{10}}

Q5

In a hydrogen like atom electron make transition from an energy level with quantum number

n

to another with quantum number

\left( {n - 1} \right)

. If

n > > 1,

the frequency of radiation emitted is proportional to :

A

{1 \over n}

B

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

C

{1 \over {{n^{{3 \over 2}}}}}

D

{1 \over {{n^3}}}

Correct Answer

Option D

Solution

\Delta

E

= 13.6{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

As

\Delta

E = h

\upsilon

$\Rightarrow$

\Delta

E $\propto$

\upsilon

$\propto$

\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

Here n1 = n - 1 and n2 = n $\therefore$

\upsilon

$\propto$

\left[ {{1 \over {{{\left( {n - 1} \right)}^2}}} - {1 \over {{n^2}}}} \right]

$\Rightarrow$

\upsilon

$\propto$

\left[ {{{2n - 1} \over {{n^2}{{\left( {n - 1} \right)}^2}}}} \right]

[ As

n > > 1,

so we can assume n - 1 = n and 2n - 1 = 2n ] $\Rightarrow$

\upsilon

$\propto$

\left[ {{{2n} \over {{n^2}{{\left( {n } \right)}^2}}}} \right]

$\propto$

\left[ {{{2n} \over {{n^4}}}} \right]

$\propto$

\left[ {{2 \over {{n^3}}}} \right]

Q6

Taking the wavelength of first Balmer line in hydrogen spectrum (n = 3 to n = 2) as 660 nm, the wavelength of the 2nd Balmer line (n = 4 to n = 2) will be :

A 642.7 nm

B 488.9 nm

C 889.2 nm

D 388.9 nm

Correct Answer

Option B

Solution

{1 \over {{\lambda _1}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right]

{1 \over {{\lambda _2}}} = R\left[ {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right]

{{5{\lambda _1}} \over {36}} = {{12{\lambda _2}} \over {4 \times 16}}

{\lambda _2} = {{5 \times 660 \times 64} \over {36 \times 12}} = 489\,nm

Q7

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :

A 194.8 nm

B 490.7 nm

C 913.3 nm

D 121.8 nm

Correct Answer

Option D

Solution

\Delta

E = 10.2 eV

{{hc} \over \lambda }

= 10.2 eV $\lambda$ =

{{hc} \over {(10.2)e}}

=

{{12400} \over {10.2}}\mathop A\limits^o

= 121.56 nm

\simeq

121.8 nm

Q8

The ratio for the speed of the electron in the 3rd orbit of He+ to the speed of the electron in the 3rd orbit of hydrogen atom will be :

A 1 : 1

B 1 : 2

C 4 : 1

D 2 : 1

Correct Answer

Option D

Solution

We know that

v \propto {Z \over n}

$\Rightarrow$ Required ratio

= {{{2 \over 3}} \over {{1 \over 3}}}

= 2:1

Q9

At a specific instant emission of radioactive compound is deflected in a magnetic field. The compound can emit:\n\n(i) electrons\n\n(ii) protons\n\n(iii)

He^{2+}

\n\n(iv) neutrons\n\nThe emission at instant can be

A

i, ii, iii

B

i, ii, iii, iv

C

iv

D

ii, iii

Correct Answer

Option A

Solution

Charged particles are deflected in magnetic field.

Q10

The atomic mass of

{ }_6 \mathrm{C}^{12}

is

12.000000 \mathrm{~u}

and that of

{ }_6 \mathrm{C}^{13}

is

13.003354 \mathrm{~u}

. The required energy to remove a neutron from

{ }_6 \mathrm{C}^{13}

, if mass of neutron is

1.008665 \mathrm{~u}

, will be :

A 62.5 MeV

B 6.25 MeV

C 4.95 MeV

D 49.5 MeV

Correct Answer

Option C

Solution

\begin{aligned} & { }_6 \mathrm{C}^{13}+\text{ Energy } \rightarrow{ }_6 \mathrm{C}^{12}+{ }_0 \mathrm{n}^1 \\ & \Delta \mathrm{m}=(12.000000+1.008665)-13.003354 \\ & =-0.00531 \mathrm{u} \\ & \therefore \text{ Energy required }=0.00531 \times 931.5 \mathrm{~MeV} \\ & =4.95 \mathrm{~MeV} \end{aligned}