Atoms and Nuclei — Page 4 | JEE Physics

Q31

If radius of the

\begin{array}{ll}{27} \\ {13} \end{array}

Al

nucleus is estimated to be

3.6

fermi then the radius of

\begin{array}{ll}{125} \\ {52} \end{array} \,Te

nucleus is estimated to be nearly

A

8

fermi

B

6

fermi

C

5

fermi

D

4

fermi

Correct Answer

Option B

Solution

KEY CONCEPT :

R = {R_0}{\left( A \right)^{1/3}}

Here A = Mass number $\therefore$

{{{R_1}} \over {{R_2}}} = {\left( {{{{A_1}} \over {{A_2}}}} \right)^{1/3}}

= {\left( {{{27} \over {125}}} \right)^{1/3}} = {3 \over 5}

{R_2} = {5 \over 3} \times 3.6 = 6

fermi

Q32

Starting with a sample of pure

{}^{66}Cu,{7 \over 8}

of it decays into

Zn

in

15

minutes. The corresponding half life is

A

15

minutes

B

10

minutes

C

7{1 \over 2}

minutes

D

5

minutes

Correct Answer

Option D

Solution

{7 \over 8}

of

Cu

decays in

15

minutes. $\therefore$

Cu

undecayed

= N = 1 - {7 \over 8} = {1 \over 8} = {\left( {{1 \over 2}} \right)^3}

$\therefore$ No. of half lifes

=3

n = {t \over T}

or

3 = {{15} \over T}

\Rightarrow T =

half life period

= {{15} \over 3} = 5\,\,

minutes

Q33

The intensity of gamma radiation from a given source is

L

. On passing through

36

mm

of lead, it is reduced to

{{\rm I} \over 8}.

The thickness of lead which will reduce the intensity to

{{\rm I} \over 2}

will be

A

9mm

B

6mm

C

12mm

D

18mm

Correct Answer

Option C

Solution

KEY CONCEPT : Intensity

I = {I_0}.{e^{ - \mu d}},

Applying logarithm on both sides,

- \mu d = \log \left( {{I \over {{I_0}}}} \right)

- \mu \times 36 = \log \left( {{{I/8} \over I}} \right).........\left( i \right)

- \mu \times d = \log \left( {{{I/2} \over I}} \right).........\left( {ii} \right)

Dividing

(i)

by

(ii),

{{36} \over d} = {{\log \left( {{1 \over 8}} \right)} \over {\log \left( {{1 \over 2}} \right)}}

= {{3\log \left( {{1 \over 2}} \right)} \over {\log \left( {{1 \over 2}} \right)}} = 3

or

\,\,\,d = {{36} \over 3} = 12\,mm

Q34

When

{}_3L{i^7}

nuclei are bombarded by protons, and the resultant nuclei are

{}_4B{e^8}

, the emitted particles will be

A alpha particles

B beta particles

C gamma photons

D neutrons

Correct Answer

Option C

Solution

{}_3^7Li + {}_1^1p \to {}_4^8Be + {}_0^0\gamma

Q35

The

'rad'

is the correct unit used to report the measurement of

A the ability of a beam of gamma ray photons to produce ions in a target

B the energy delivered by radiation to a target

C the biological effect of radiation

D the rate of decay of radioactive source

Correct Answer

Option C

Solution

The risk posed to a human being by any radiation exposure depends partly upon the absorbed dose, the amount of energy absorbed per gram of tissue.

Absorbed dose is expressed in rad.

A rad is equal to

100

ergs

of energy absorbed by

1

gram of tissue. The more modern, internationally adopted unit is the gray (named after the English medical physicist

L.

H.

Gray); one gray equals

100

rad.

Q36

An alpha nucleus of energy

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

bombards a heavy nuclear target of charge

Ze

. Then the distance of closest approach for the alpha nucleus will be proportional to

A

{v^2}

B

{1 \over m}

C

{1 \over {{v^2}}}

D

{1 \over {Ze}}

Correct Answer

Option C

Solution

Work done to stop the $\alpha$ particle is equal to

K.E.

$\therefore$

qV = {1 \over 2}m{v^2} \Rightarrow q \times {{K\left( {Ze} \right)} \over r} = {1 \over 2}m{v^2}

\Rightarrow r = {{2\left( {2e} \right)K\left( {Ze} \right)} \over {m{V^2}}} = {{4KZ{e^2}} \over {m{v^2}}}

\Rightarrow r \propto {1 \over {{v^2}}}

and

r \propto {1 \over m}.

Q37

If

{M_O}

is the mass of an oxygen isotope

{}_8{O^{17}}

,

{M_p}

and

{M_N}

are the masses of a proton and neutron respectively, the nuclear binding energy of the isotope is

A

\left( {{M_O} - 17{M_N}} \right){C^2}

B

\left( {{M_O} - 8{M_P}} \right){C^2}

C

\left( {{M_O} - 8{M_P} - 9{M_N}} \right){C^2}

D

{{M_O}{c^2}}

Correct Answer

Option C

Solution

Binding energy

= \left[ {Z{M_p} + \left( {A - Z} \right){M_N} - M} \right]{c^2}

= \left[ {8{M_p} + \left( {17 - 8} \right){M_N} - M} \right]{c^2}

= \left[ {8{M_p} + 9{M_N} - M} \right]{c^2}

= \left[ {8{M_p} + 9{M_N} - {M_O}} \right]{c^2}

Q38

In gamma ray emission from a nucleus

A only the proton number changes

B both the neutron number and the proton number change

C there is no change in the proton number and the neutron number

D only the neutron number changes

Correct Answer

Option C

Solution

There is no change in the proton number and the neutron number as the $\gamma$ - emission takes place as a result of excitation or de-excitation of nuclei. $\gamma$ -rays have no charge or mass.

Q39

The half-life period of a ratio-active element

X

is same as the mean life time of another ratio-active element

Y.

Initially they have the same number of atoms. Then

A

X

and

Y

decay at same rate always

B

X

will decay faster than

Y

C

Y

will decay faster than

X

D

X

and

Y

have same decay rate initially

Correct Answer

Option C

Solution

According to question, Half life of

X,\,{T_{1/2}} = {\tau _{av}},\,\,\,

average life of

Y

\Rightarrow {{0.693} \over {{\lambda _X}}} = {1 \over {{\lambda _Y}}} \Rightarrow {\lambda _X} = \left( {0.693} \right).{\lambda _Y}

$\therefore$

{\lambda _X} < {\lambda _Y}.

Now, the rate of decay is given by

- {{dN} \over {dt}} = \lambda N

$\therefore$

Y

will decay faster than

X.

[ as

N

is same ]

Q40

Which of the following transitions in hydrogen atoms emit photons of highest frequency ?

A

n = 1

to

n=2

B

n = 2

to

n=6

C

n = 6

to

n=2

D

n = 2

to

n=1

Correct Answer

Option D

Solution

We have no find the frequency of emitted photons.

For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options

(c)

and

(d)

. Frequency is given by

hv = - 13.6\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)

For transition from

n=6

to

n=2,

{v_1} = {{ - 13.6} \over h}\left( {{1 \over {{6^2}}} - {1 \over {{2^2}}}} \right)

= {2 \over 9} \times \left( {{{13.6} \over h}} \right)

For transition from

n=2

to

n=1,

{v_2} = {{ - 13.6} \over h}\left( {{1 \over {{2^2}}} - {1 \over {{1^2}}}} \right)

= {3 \over 4} \times \left( {{{13.6} \over h}} \right).

$\therefore$

{v_1} > {v_2}