Atoms and Nuclei — Page 8 | NEET Physics

Q71

In a Rutherford scattering experiment when a projectile of charge z 1 and mass M 1 approaches a target nucleus of charge z 2 and mass M 2 , the distance of closest approach is r 0 . The energy of the projectile is

A directly proportional to z 1 z 2

B inversely proportional to z 1

C directly proportional to mass M 1

D directly proportional to M 1

\times

M 2

Correct Answer

Option A

Solution

Now energy of the projectile : =

{1 \over {4\pi {\varepsilon _0}}}{{{z_1}{z_2}} \over r}

Therefore energy $\propto$ z 1 z 2

Q72

If M(A; Z), M p and M n denote the masses of the nucleus

{}_Z^AX,

proton and neutron respectively in units of u (1 u = 931.5 MeV/c 2 ) and BE represents its bonding energy in MeV, then

A M(A, Z) = ZM p + (A

-

Z)M n

-

BE

B M(A, Z) = ZM p + (A

-

Z)M n + BE/c 2

C M(A, Z) = ZM p + (A

-

Z)M n

-

BE/c 2

D M(A, Z) = ZM p + (A

-

Z)M n + BE

Correct Answer

Option C

Solution

Mass defect = ZM p + (A –Z)M n – M(A, Z) $\Rightarrow$

{{B.E} \over {{c^2}}}

= ZM p + (A –Z)M n – M(A, Z) $\therefore$ M(A, Z) = ZM p + (A $-$ Z)M n $-$

{{B.E} \over {{c^2}}}

Q73

Two radioactive materials X 1 and X 2 have decay constants

5\lambda

and

\lambda

respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X 1 to that X 2 will be 1/e after a time

A 1/4

\lambda

B e/

\lambda

C

\lambda

D

{1 \over 2}\lambda

Correct Answer

Option A

Solution

X 1 = N 0 e –5 $\lambda$ t X 2 = N 0 e – $\lambda$ t Given

{{{X_1}} \over {{X_2}}}

=

{1 \over e}

$\Rightarrow$

{{{N_0}{e^{ - 5\lambda t}}} \over {{N_0}{e^{ - \lambda t}}}}

=

{1 \over e}

$\Rightarrow$ e –4 $\lambda$ t = e -1 $\Rightarrow$ t =

{1 \over {4\lambda }}

Q74

The ground state energy of hydrogen atom is

-

13.6 eV. When its electron is in the first excited state, its excitation energy is

A 10.2 eV

B 0

C 3.4 eV

D 6.8 eV

Correct Answer

Option A

Solution

E n =

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

E 1 = -13.6 eV E 2 =

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

= -3.4 eV $\therefore$

\Delta

E = ( – 3.4) – ( – 13.6) = 10.2 eV.

Q75

Two nuclei have their mass numbers in the ratio of 1 : 3. The ratio of their nuclear densities would be

A

{\left( 3 \right)^{1/3}}:1

B 1 : 1

C 1 : 3

D 3 : 1

Correct Answer

Option B

Solution

A 1 : A 2 = 1 : 3 Their radii will be in the ratio

{R_0}A_1^{1/3}

:

{R_0}A_2^{1/3}

= 1 : 3 1/3 $\therefore$

{\rho _{{A_1}}}

:

{\rho _{{A_2}}}

=

{{{A_1}} \over {{4 \over 3}\pi R_1^3}}

:

{{{A_2}} \over {{4 \over 3}\pi R_2^3}}

=

{1 \over {{1^3}}}:{3 \over {{{\left( {{3^{1/3}}} \right)}^3}}}

= 1 : 1

Q76

The total energy of electron in the ground state of hydrogen atom is

-

13.6 eV. The kinetic energy of an electron in the first excited state is

A 6.8 eV

B 13.6 eV

C 1.7 eV

D 3.4 eV.

Correct Answer

Option D

Solution

E n =

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

E 1 = -13.6 eV E 2 =

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

= -3.4 eV Kinetic energy of an electron in the first excited state is K = –E 2 = 3.4 eV.

Q77

If the nucleus

{}_{13}^{27}Al

has a nuclear radius of about 3.6 fm, them

{}_{32}^{125}Te

would have its radius approximately as

A 9.6 fm

B 12.0 fm

C 4.8 fm

D 6.0 fm

Correct Answer

Option D

Solution

A nucleus of mass number A has radius R = R 0 A 1/3 , where R 0 = 1.2 × 10 –15 m and A = mass number For

{}_{13}^{27}Al

, R 1 = R 0 (27) 1/3 = 3R 0 For

{}_{32}^{125}Te

, R 2 = R 0 (125) 1/3 = 5R 0 $\therefore$

{{{R_2}} \over {{R_1}}} = {{5{R_0}} \over {3{R_0}}}

$\Rightarrow$

{R_2} = {5 \over 3} \times {R_1}

=

{5 \over 3} \times 3.6

= 6.0 fm

Q78

A nucleus

{}_Z^AX

has mass represented by M(A, Z). If M p and M n denote the mass of proton and neutron respectively and B.E. the binding energy in MeV, then

A B.E. = [ZM p + (A

-

Z)M n

-

M(A, Z)]c 2

B B.E. = [ZM p + AM p

-

M(A, Z)]c 2

C B.E. = M(A, Z)

-

ZM p

-

(A

-

Z)M n

D B.E. = [M(A, Z)

-

ZM p

-

(A

-

Z)M n ]c 2

Correct Answer

Option A

Solution

The difference in mass of a nucleus and its constituents,

\Delta

M, is called the mass defect and is given by

\Delta

M = [ZM p + (A $-$ Z)M n ] - M(A, Z) binding energy =

\Delta

Mc 2 = [ZM p + (A $-$ Z)M n $-$ M(A, Z)]c 2

Q79

In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential

V

and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio

\left( {{{ch\arg e\,\,on\,\,\,the\,\,ion\,\,} \over {mass\,\,of\,\,the\,\,ion}}} \right)

will be proportional to

A 1/R 2

B R 2

C R

D 1/R

Correct Answer

Option A

Solution

In mass spectrometer, when ions are accelerated through potential V,

{1 \over 2}m{v^2} = qV

....(1) As the magnetic field curves the path of the ions in a semicircular orbit. $\therefore$ Bqv =

{{m{v^2}} \over R}

$\Rightarrow$ v =

{{BqR} \over m}

.....(2) Now by substituting (ii) in (i),

{1 \over 2}m{\left( {{{BqR} \over m}} \right)^2} = q

$\Rightarrow$

{q \over m} = {{2V} \over {{B^2}{R^2}}}

As V and B are constants, $\therefore$

{q \over m}

$\propto$

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

Q80

In a radioactive decay process, the negatively charged emitted

\beta

-particles are

A the electrons produced as a result of the decay of neutrons inside the nucleus

B the electrons produced as a result of collisions between atoms

C the electrons orbitting around the nucleus

D the electrons present inside the nucleus

Correct Answer

Option A

Solution

In beta minus decay ( $\beta$ – ), a neutron is transformed into a proton, and an electron is emitted from the nucleus along with antineutrino. n = p + e $-$ +

\overline \nu