Atoms and Nuclei — Page 7 | NEET Physics

Q61

The half life of a radioactive isotope X is 50 years. It decays to another element Y which is stable. The two elements X and Y were found to be in the ratio of 1 : 15 in a sample of a given rock. The age of the rock was estimated to be

A 150 years

B 200 years

C 250 years

D 100 years

Correct Answer

Option B

Solution

Initial number of atoms of X is N 0 and Initial number of atoms of Y is 0 Number of atoms after time t, for X is N and for Y is N 0 - N According to the question,

{N \over {{N_0} - N}}

=

{1 \over 15}

$\Rightarrow$

{N \over {{N_0}}} = {1 \over 16}

As

{N \over {{N_0}}} = {\left( {{1 \over 2}} \right)^n}

where n is the no. of half lives $\therefore$

{\left( {{1 \over 2}} \right)^n} = {1 \over 16}

$\Rightarrow$ n = 4 t = nT 1/2 = 4 × 50 = 200 years Hence, the age of rock is 200 years.

Q62

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is

A 3

B 4

C 1

D 2

Correct Answer

Option D

Solution

The wavelength of the first line of lyman series for hydrogen atom is

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

The wavelength of the second line of Balmer series for hydrogen like ion is

{1 \over {\lambda '}} = R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)

According to question $\lambda$ = $\lambda$ '

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

=

R{Z^2}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)

$\Rightarrow$

{3 \over 4}

=

{{3{Z^2}} \over {16}}

$\Rightarrow$ Z = 2

Q63

The decay constant of a radio isotope is

\lambda

. If A 1 and A 2 are its activities at times t 1 and t 2 respectively, the number of nuclei which have decayed during the time (t 1

-

t 2 )

A A 1 t 1

-

A 2 t 2

B A 1

-

A 2

C (A 1

-

A 2 )/

\lambda

D

\lambda ({A_1} - {A_2})

Correct Answer

Option C

Solution

A 1 = $\lambda$ N 1 at time t 1 A 2 = $\lambda$ N 2 at time t 2 Therefore, number of nuclei decayed during time interval (t 1 – t 2 ) is N 1 - N 2 =

{{\left[ {{A_1} - {A_2}} \right]} \over \lambda }

Q64

The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

A 23.6 MeV

B 2.2 MeV

C 28.0 MeV

D 30.2 MeV

Correct Answer

Option A

Solution

1 H 2 + 1 H 2 $\to$ 2 He 4 +

\Delta

E The binding energy per nucleon of a deuteron = 1.1 MeV $\therefore$ Total binding energy = 2 × 1.1 = 2.2 MeV The binding energy per nucleon of a helium nuclei = 7 MeV $\therefore$ Total binding energy = 4 × 7 = 28 MeV Hence, energy released

\Delta

E = (28 – 2 × 2.2) = 23.6 MeV

Q65

The activity of a radioactive sample is measured as N 0 counts per minute at t = 0 and N 0 /e counts per minute at t = 5 minutes. The time (in minutes) at which the activity reduces to half its value is

A

{\log _e}{2 \over 5}

B

{5 \over {{{\log }_e}2}}

C 5log 10 2

D 5log e 2

Correct Answer

Option D

Solution

Now N = N 0 e – $\lambda$ t $\Rightarrow$

{{{{N_0}} \over e}}

= N 0 e –5t $\Rightarrow$ $\lambda$ =

{1 \over 5}

per minute Further,

{{{N_0}} \over 2}

= N 0 e –λ(t) t =

{1 \over \lambda }\ln 2

= 5log e 2

Q66

The energy of a hydrogen atom in the ground state is

-

13.6 eV. The energy of a He + ion in the first excited state will be

A

-

13.6 eV

B

-

27.2 eV

C

-

54.4 eV

D

-

6.8 eV

Correct Answer

Option A

Solution

Energy of a H-like atom in it's n th state is given by, E n =

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

For, first excited state of He + , n = 2, Z = 2 $\therefore$ E He + =

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

= -13.6 eV

Q67

The mass of a

{}_3^7Li

Li nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of

{}_3^7Li

nucleus is nearly

A 46 MeV

B 5.6 MeV

C 3.9 MeV

D 23 MeV

Correct Answer

Option B

Solution

For

{}_3^7Li

nucleus, Mass defect,

\Delta

M = 0.042 u $\therefore$ 1 u = 931.5 MeV/c 2 $\therefore$

\Delta

M = 0.042 × 931.5 MeV/c 2 = 39.1 MeV/c 2 Number of nucleons in

{}_3^7Li

is 7. Binding energy per nucleon, E bn =

{{39.1} \over 7}

= 5.6 MeV

Q68

An alpha nucleus of energy

{1 \over 2}

mv 2 bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to

A

{1 \over {Ze}}

B v 2

C

{1 \over m}

D

{1 \over {{v_4}}}

Correct Answer

Option C

Solution

Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus.

That is,

{1 \over 2}m{v^2} = {1 \over {4\pi {\varepsilon _0}}} \times {{\left( {Ze} \right) \times \left( {2e} \right)} \over {{d_{in}}}}

$\therefore$ d min $\propto$

{1 \over m}

Q69

The number of beta particles emitted by a radioactive substance is twice the number of alpha particles emitted by it. The resulting daughter is an

A isomer of parent

B isotone of parent

C isotope of parent

D isobar of parent

Correct Answer

Option C

Solution

{}_Z^AX\overset{{2{\beta ^ - }}}\longrightarrow {}_{Z + 2}^A{Y_1}\overset{\alpha}\longrightarrow {}_Z^{A + 4}{Y_2}

The resultant daughter is an isotope of the original parent nucleus.

Q70

The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths, Maximum wavelength of emitted radiation corresponds to the transition between

A n = 3 to n = 1 states

B n = 2 to n = 1 states

C n = 4 to n = 3 states

D n = 3 to n = 2 states

Correct Answer

Option C

Solution

n(n – 1)/2 = 6 $\Rightarrow$ n 2 – n –12 = 0 $\Rightarrow$ (n – 4) (n + 3) = 0 or n = 4 So, n = 4 to n = 3 states.