Atoms and Nuclei — Page 10 | NEET Physics

Q91

In the reaction

{}_1^2

H +

{}_1^3

H

\to

{}_2^4

He +

{}_0^1

n, if the binding energies of

{}_1^2

H,

{}_1^3

H and

{}_2^4

He are respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is

A a + b + c

B a + b

-

c

C c

-

a

-

b

D c + a

-

b

Correct Answer

Option C

Solution

Energy released in given reaction = BE of products – BE of reactants = c – (a+ b) = c – a – b

Q92

The half life of radian is about 1600 years. Of 100 g of radium existing now, 25 g will remain unchanged after

A 4800 years

B 6400 years

C 2400 years

D 3200 years

Correct Answer

Option D

Solution

N = N 0

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

$\Rightarrow$

{{25} \over {100}}

=

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

$\Rightarrow$ n = 2 The total time in which radium change to 25 g is = 2 × 1600 = 3200 yr.

Q93

The Bohr model of atoms

A Assumes that the angular momentum of electrons is quantized.

B Uses Einstein's photoelectric equation.

C Predicts continuous emission spectra fror atoms.

D Predicts the same emission spectra for all types of atoms.

Correct Answer

Option A

Solution

Bohr model of atoms assumes that the angular momentum of electrons is quantised.

Q94

If in a nuclear fusion process the masses of the fusing nuclei be m 1 and m 2 and the mass of the resultant nucleus be m 3 , then

A m 3 = m 1 + m 2

B m 3 =

\left| {{m_1} - {m_2}} \right|

C m 3 < (m 1 + m 2 )

D m 3 > (m 1 + m 2 )

Correct Answer

Option C

Solution

In nuclear fusion the mass of end product or resultant is always less than the sum of initial product, the rest is liberated in the form of energy, like in Sun energy is liberated due to fusion of two hydrogen atoms.

As m 1 + m 2 = m 3 + E Since E = (m 1 + m 2 – m 3 ) × c 2

Q95

A nucleus represented by the symbol

{}_Z^AX

has

A Z neutrons and A

-

Z protons

B Z protons and A

-

Z neutrons

C Z protons and A neutrons

D A protons and Z

-

A neutrons

Correct Answer

Option B

Solution

Z is number of protons and A is the total number of protons and neutrons.

Q96

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}}}

Q97

The mass number of a nucleus is

A always less than its atomic number

B always more than its atomic number

C sometimes equal to its atomic number

D sometimes less than and sometimes more than its atomic number

Correct Answer

Option C

Solution

Mass number = atomic number + no. of neutrons For hydrogen, number of neutrons = 0 So, mass number = Atomic number.

Hence mass number is sometimes equal to atomic number.

Q98

The mass of proton is 1.0073 u and that of neutron is 1.0087 u (u = atomic mass unit). The binding energy of

{}_2^4

He is (Given helium nucleus mass

\approx

4.0015 u.)

A 0.0305 J

B 0.0305 erg

C 28.4 MeV

D 0.061 u

Correct Answer

Option C

Solution

Mass defect(

\Delta

m) = 2M P + 2M N – M He $\Rightarrow$

\Delta

m = 2 × 1.0074 + 2 × 1.0087 – 4.0015 $\Rightarrow$

\Delta

m = 0.0307 Now as per relation, E =

\Delta

m × 931 MeV = 0.0307 × 931 = 28.4 MeV

Q99

A sample of radioactive element has a mass of 10 g at an instant t = 0. The approximate mass of this element in the sample after two mean lives is

A 1.35 g

B 2.50 g

C 3.70 g

D 6.30 g

Correct Answer

Option A

Solution

Using the relation for mean life. Given : t = 2t = 2

\left( {{1 \over \lambda }} \right)

Then from M = M 0 e - $\lambda$ t = 10e - $\lambda$

\left( {{1 \over \lambda }} \right)

=

10{\left( {{1 \over e}} \right)^2}

= 1.35 g

Q100

The volume occupied by an atom is greater than the volume of the nucleus by a factor of about

A 10 1

B 10 5

C 10 10

D 10 15

Correct Answer

Option D

Solution

As V $\propto$ r 3 $\Rightarrow$ V atom =

{{r_{atom}^3} \over {r_{nucleus}^3}}

$\times$ V nucleus =

{{{{\left( {{{10}^{ - 10}}} \right)}^3}} \over {{{\left( {{{10}^{ - 15}}} \right)}^3}}}

$\times$ V nucleus = 10 15 $\times$ V nucleus