Atoms and Nuclei — Page 6 | JEE Physics

Q51

Hydrogen

\left( {{}_1{H^1}} \right)

, Deuterium

\left( {{}_1{H^2}} \right)

, singly ionised Helium

{\left( {{}_2H{e^4}} \right)^ + }

and doubly ionised lithium

{\left( {{}_3L{i^6}} \right)^{ + + }}

all have one electron around the nucleus. Consider an electron transition from

n=2

to

n=1.

If the wavelengths of emitted radiation are

{\lambda _1},{\lambda _2},{\lambda _3}

and

{\lambda _4}

respectively then approximately which one of the following is correct?

A

4{\lambda _1} = 2{\lambda _2} = 2{\lambda _3} = {\lambda _4}

B

{\lambda _1} = 2{\lambda _2} = 2{\lambda _3} = {\lambda _4}

C

{\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}

D

{\lambda _1} = 2{\lambda _2} = 3{\lambda _3} = 4{\lambda _4}

Correct Answer

Option C

Solution

Wave number

{1 \over \lambda } = R{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {{n^2}}}} \right]

\Rightarrow \lambda \propto {1 \over {{Z^2}}}

By question

n=1

and

{n_1} = 2

Then,

{\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}

Q52

As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion :

A kinetic energy decreases, potential energy increases but total energy remains same

B kinetic energy and total energy decrease but potential energy increases

C its kinetic energy increases but potential energy and total energy decrease

D kinetic energy, potential energy and total energy decrease

Correct Answer

Option C

Solution

U = - K{{z{e^2}} \over r};\,\,T.E = {k \over 2}{{ze^2} \over r}

K.E = {k \over 2}{{z{e^2}} \over r}.

Here

r

decreases

Q53

A He+ ion is in its first excited state. Its ionization energy is :-

A 13.60 eV

B 6.04 eV

C 48.36 eV

D 54.40 eV

Correct Answer

Option A

Solution

T.E. = - \left( {13.6} \right)\left( {{{{Z^2}} \over {{n^2}}}} \right)eV

z = n = 2 $\Rightarrow$ Ionisation energy = –T.E. = 13.6 eV

Q54

Half-lives of two radioactive elements

A

and

B

are

20

minutes and

40

minutes, respectively. Initially, the samples have equal number of nuclei. After

80

minutes, the ratio of decayed number of

A

and

B

nuclei will be:

A

1:4

B

5:4

C

1:16

D

4:1

Correct Answer

Option B

Solution

For

{A_{t{{1}/{2}}}} = 20\,\,

min,

t=80

min, number of half lifes

n=4

$\therefore$ Nuclei remaining

= {{{N_0}} \over {{2^4}}}.

Therefore nuclei decayed

= {N_0} - {{{N_0}} \over {{2^4}}}

For

{B_{t{{1}/{2}}}} = 40\,\,

min,

t=80

min, number of half lifes

n=4

$\therefore$ Nuclei remaining

= {{{N_0}} \over {{2^2}}}.

Therefore nuclei decayed

= {N_0} - {{{N_0}} \over {{2^2}}}

$\therefore$ Required ratio

= {{{N_0} - {{{N_0}} \over {{2^4}}}} \over {{N_0} - {{{N_0}} \over {{2^2}}}}}

= {{1 - {1 \over {16}}} \over {1 - {1 \over 4}}}

= {{15} \over {16}} \times {4 \over 3} = {5 \over 4}

Q55

Imagine that a reactor converts all given mass into energy and that it operates at a power level of 109 watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is 3×108 m/s)

A 0.96 gm

B 0.8 gm

C 4

\times

10

-

2 gm

D 6.6

\times

10

-

5 gm

Correct Answer

Option C

Solution

The power can be calculated by the relation

P = {E \over {\Delta t}} = {{\Delta m{c^2}} \over {\Delta t}}

...... (1) Therefore, from Eq. (1), the mass of the fuel consumed per hour in the reactor is

{{\Delta m} \over {\Delta t}} = {P \over {{c^2}}} = {{{{10}^9}} \over {{{(3 \times {{10}^8})}^2}}} = 4 \times {10^{ - 12}}

g

Q56

The acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is :

A

{{{h^2}} \over {{\pi ^2}{m^2}{r^3}}}

B

{{{h^2}} \over {{8\pi ^2}{m^2}{r^3}}}

C

{{{h^2}} \over {{4\pi ^2}{m^2}{r^3}}}

D

{{{h^2}} \over {{4\pi }{m^2}{r^3}}}

Correct Answer

Option C

Solution

The speedy of the particle in the orbit of an atom is

v = {{{I^2}} \over {2h{\varepsilon _0}}}

We have the radius of the first orbit is

r = {{{h^2}{\varepsilon _0}} \over {\pi m{e^2}}}

\Rightarrow {\varepsilon _0} = {{r\pi m{e^2}} \over {{h^2}}}

Therefore, the acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is

{{{v^2}} \over r} = {{{I^6}\pi m} \over {4{h^2}\varepsilon _0^3}} = {{{h^2}} \over {4{r^3}{\pi ^2}{m^2}}}

Q57

According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to : (n = principal quantum number)

A

{n^{ - 4}}

B

{n^{ - 5}}

C n

-

3

D n

-

2

Correct Answer

Option B

Solution

Magnetic field at the center of neucleus of H-atom B =

{{{\mu _0}I} \over {2{r_n}}}

Radius of nth orbital, rn =

{{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}

\therefore\,\,\,

rn $\propto$ n2 velocity of electron in nth orbital,

\upsilon

n =

\left( {{{{e^2}h} \over {2{\varepsilon _0}}}} \right){Z \over n}

\therefore\,\,\,

\upsilon

n $\propto$ n $-$ 1 I =

{q \over t}

=

{e \over {{{2\pi {r_n}} \over {{\upsilon _n}}}}}

=

{{e{\upsilon _n}} \over {2\pi {r_n}}}

\therefore\,\,\,

B =

{{{\mu _0}.\left( {{{e{\upsilon _n}} \over {2\pi {r_n}}}} \right)} \over {2{r_n}}}

=

{{{\mu _0}\,e{\upsilon _n}} \over {4\pi r_n^2}}

\therefore\,\,\,

B $\propto$

{{{\upsilon _n}} \over {r_n^2}}

$\Rightarrow$

\,\,\,

B $\propto$

{{{n^{ - 1}}} \over {{n^4}}}

$\Rightarrow$

\,\,\,

B $\propto$ n $-$ 5

Q58

Two radioactive substances A and B have decay constants 5

\lambda

and

\lambda

respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the number of nuclei to become

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

will be :

A

{2 \over \lambda }

B

{1 \over {4\lambda }}

C

{1 \over {2\lambda }}

D

{1 \over {\lambda }}

Correct Answer

Option C

Solution

Nx(at t) = N0e–5 $\lambda$ t Ny(at t) = N0e– $\lambda$ t

{{{N_x}} \over {{N_y}}} = {1 \over {{e^2}}} = {e^{ - 4\lambda t}}

\Rightarrow 4\lambda t = 2

\Rightarrow t = {2 \over {4\lambda }} = \left( {{1 \over {2\lambda }}} \right)

Q59

A radioactive nucleus A with a half life T, decays into a nucleus B. At t = 0, there is no nucleus B. At sometime t, the ratio of the number of B to that of A is 0.3. Then, t is given by :

A

t = {T \over {\log (1.3)}}

B

t = T\log (1.3)

C

t = {T \over 2}{{\log 2} \over {\log 1.3}}

D

t = T{{\log 1.3} \over {\log 2}}

Correct Answer

Option D

Solution

Let initially there are total N0 number of nuclei. At time t

{{{N_B}} \over {{N_A}}}

= 0.3 (given) $\Rightarrow$

{{N_B}}

= 0.3

{{N_A}}

N0 = NA + NB = NA + 0.3NA $\therefore$ NA =

{{{N_0}} \over {1.3}}

As we know Nt = N0 e– $\lambda$ t $\Rightarrow$

{{{N_0}} \over {1.3}}

= N0 e– $\lambda$ t $\Rightarrow$

{1 \over {1.3}}

= e– $\lambda$ t $\Rightarrow$ ln(1.3) = $\lambda$ t $\Rightarrow$ t =

{{\ln \left( {1.3} \right)} \over \lambda }

If T is half-life, then $\lambda$ =

{{{{\ln} } 2} \over T}

$\Rightarrow$ t =

{{\ln \left( {1.3} \right)} \over {{{\ln 2} \over T}}}

=

{{\ln \left( {1.3} \right)} \over {\ln 2}}T

Q60

A solution containing active cobalt

{^{60}_{27}}Co

having activity of

0.8

\mu Ci

and decay constant

\lambda

is injected in an animal's body. If

1\,c{m^3}

of blood is drawn from the animal's body after

10

hrs of injection, the activity found was

300

decays per minute What is the volume of blood that is flowing in the body ?

\left( {\,\,Ci = 3.7 \times {{10}^{10}}\,} \right.

decays per second and at

t=10

hrs

\left. {{e^{ - \lambda t}} = 0.84} \right)

A

6

liters

B

7

liters

C

4

liters

D

5

liters

Correct Answer

Option D

Solution

Initial activity, No = 0.8 $\mu$ Ci Activity at time t, N = N0 e $-$ $\lambda$ t Activity in 1 cm3 blood after 10 hr, n = 300 decays per minute =

{{300} \over {60}}\,dps

= 5 dps. Activity in whole blood after to hr = N0 e $-$ $\lambda$ $\times$ 10

\therefore\,\,\,\,

Volume of the total blood =

{{{N_0}{e^{ - 10\lambda }}} \over n}

=

{{0.8 \times {{10}^{ - 7}} \times 3.7 \times {{10}^{10}} \times 0.84} \over 5}

= 4.97 $\times$ 103 cm3 = 4.97 litres

\simeq

5 liters.