Atoms and Nuclei — Page 5 | NEET Physics

Q41

The binding energy per nucleon of and nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction

{}_3^7Li + {}_1^1H \to {}_2^4He + _2^4He + Q

the value of energy Q released is

A 19.6 MeV

B

-

2.4 MeV

C 8.4 MeV

D 17.3 MeV

Correct Answer

Option D

Solution

Given: Binding energy per nucleon of 3 Li 7 and 2 He 4 nuclei are 5.60 MeV and 7.06 MeV Energy released = 7.06 × 8 – 5.60 × 7 = 17.3 MeV

Q42

A radioactive X with a half life 1.4

\times

10 9 years decays to Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1 : 7. The age of the rock is

A 1.96

\times

10 9 years

B 3.92

\times

10 9 years

C 4.20

\times

10 9 years

D 8.40

\times

10 9 years

Correct Answer

Option C

Solution

Using N = N 0 e –λt Now 1 = 8e –λt or,

{1 \over 8}

= e – (ln 2/T)t or,

{\left( {{1 \over 2}} \right)^3}

= (2) –t/T or, t/T = 3 Now time, t = 3 × 1.4 × 10 9 = 4.2 × 10 9 years

Q43

\alpha

-particles,

\beta

-particles and

\gamma

-rays are all having same energy. Their penetrating power in a given medium in increasing order will be

A

\gamma

,

\alpha

,

\beta

B

\alpha

,

\beta

,

\gamma

C

\beta

,

\alpha

,

\gamma

D

\beta

,

\gamma

,

\alpha

Correct Answer

Option B

Solution

For a given energy, $\gamma$ -rays has highest penetrating power and $\alpha$ -particles has least penetrating power.

Q44

An electron in hydrogen atom makes a transition n 1

\to

n 2 where n 1 and n 2 are principal quantum numbers of the two states . Assuming Bohr's model to be valid, the time period of the electron in the initial state is eight times that in the final state. The possible values of n 1 and n 2 are

A n 1 = 6 and n 2 = 2

B n 1 = 8 and n 2 = 1

C n 1 = 8 and n 2 = 2

D n 1 = 4 and n 2 = 2

Correct Answer

Option D

Solution

As T $\propto$ n 3 $\therefore$

{{{T_1}} \over {{T_2}}} = {{8{T_2}} \over {{T_2}}}

=

{\left( {{{{n_1}} \over {{n_2}}}} \right)^3}

Hence, n 1 = 2n 2

Q45

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen spectrum is

A

{7 \over {29}}

B

{9 \over {31}}

C

{5 \over {27}}

D

{3 \over {23}}

Correct Answer

Option C

Solution

We see that wavelength of Lyman series, n f = 3, n i = 4

{1 \over {{\lambda _L}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right]

=

{{7R} \over {144}}

We see that wavelength of Balmer series : n f = 2, n i = 3

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

=

{{5R} \over {36}}

Now ratio of longest wavelengths corresponds to Lyman and Balmer series:

{{{\lambda _L}} \over {{\lambda _B}}} = {5 \over {36}} \times {{144} \over 7}

=

{5 \over {27}}

Q46

A certain mass of Hydrogen is changed to Helium by the process of fusion. The mass defect in fusion reaction is 0.02866 u. The energy liberated per u is (given 1 u = 931 MeV)

A 6.675 MeV

B 13.35 MeV

C 2.67 MeV

D 26.7 MeV

Correct Answer

Option A

Solution

Mass defect

\Delta

m = 0.02866 a.m.u. Energy = 0.02866 × 931 = 26.7 MeV As 1 H 2 + 1 H 2 $\to$ 2 He 4 Energy liberated per a.m.u =

{{13.35} \over 2}

MeV = 6.675 MeV

Q47

The half life of a radioactive isotope 'X' is 20 years. It decays to another element 'Y' which is stable. The two elements 'X' and 'Y' were found to be in the ratio 1 : 7 in a sample of a given rock. The age of the rock is estimated to be

A 80 years

B 100 years

C 40 years

D 60 years

Correct Answer

Option D

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 7}

$\Rightarrow$

{N \over {{N_0}}} = {1 \over 8}

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 8}

$\Rightarrow$ n = 3 t = nT 1/2 = 3 × 20 = 60 years Hence, the age of rock is 60 years.

Q48

The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from

A 2

\to

1

B 3

\to

2

C 4

\to

2

D 4

\to

3

Correct Answer

Option D

Solution

The frequency of the transition v $\propto$

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

when n = 1, 2, 3.

Q49

The half life of a radioactive nucleus is 50 days. The time invertal (t 2

-

t 1 ) between the time t 2 when

{2 \over 3}

of it has decayed and the time t 1 when

{1 \over 3}

of it had decayed is

A 30 days

B 50 days

C 60 days

D 15 days

Correct Answer

Option B

Solution

At time t 2 ,

{2 \over 3}

of the sample had decayed $\therefore$ N =

{1 \over 3}

N 0 $\therefore$

{1 \over 3}

N 0 = N 0

{e^{ - \lambda {t_2}}}

......(1) At time t 1 ,

{1 \over 3}

of the sample had decayed $\therefore$ N =

{2 \over 3}

N 0 $\therefore$

{2 \over 3}

N 0 = N 0

{e^{ - \lambda {t_1}}}

......(2) Divide (i) by (ii), we get

{1 \over 2} = {{{e^{ - \lambda {t_2}}}} \over {{e^{ - \lambda {t_1}}}}}

$\Rightarrow$

{1 \over 2} = {e^{ - \lambda \left( {{t_2} - {t_1}} \right)}}

$\Rightarrow$

{\lambda \left( {{t_2} - {t_1}} \right)}

= ln 2 $\Rightarrow$

{\left( {{t_2} - {t_1}} \right)}

=

{{\ln 2} \over \lambda }

=

{{\ln 2} \over {{{\ln 2} \over {{T_{1/2}}}}}}

= T 1/2 = 50 days

Q50

A mixture consists of two radioactive materials A 1 and A 2 with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of A 1 and 160 g of A 2 . The amount of the two in the mixture will become equal after

A 60 s

B 80 s

C 20 s

D 40 s

Correct Answer

Option D

Solution

Let, the amount of the two in the mixture will become equal after t years.

The amount of A 1 , which remains after t years N 1 =

{{{N_{01}}} \over {{{\left( 2 \right)}^{t/20}}}}

The amount of A 2 , which remains, after t years N 2 =

{{{N_{02}}} \over {{{\left( 2 \right)}^{t/10}}}}

According to the question N 1 = N 2 $\Rightarrow$

{{40} \over {{{\left( 2 \right)}^{t/20}}}} = {{160} \over {{{\left( 2 \right)}^{t/10}}}}

$\Rightarrow$

{{{\left( 2 \right)}^{t/20}}}

=

{{{\left( 2 \right)}^{\left( {{t \over {10}} - 2} \right)}}}

$\Rightarrow$

{{t \over {20}} = \left( {{t \over {10}} - 2} \right)}

$\Rightarrow$ t = 40 s