Atoms and Nuclei — Page 6 | NEET Physics

Q51

An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be

A

{{24hR} \over {25m}}

B

{{25hR} \over {24m}}

C

{{25m} \over {24hR}}

D

{{24m} \over {25hR}}

Correct Answer

Option A

Solution

We know,

{1 \over \lambda } = R{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)

where R = Rydberg constant, Z = atomic number Here, n 1 = 1, n 2 = 5 $\therefore$

{1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {{5^2}}}} \right)

=

{{24R} \over {25}}

According to conservation of linear momentum, we get Momentum of photon = Momentum of atom $\Rightarrow$

{h \over \lambda }

= mv $\Rightarrow$ v =

{h \over {m\lambda }}

=

{h \over m}\left( {{{24R} \over {25}}} \right)

=

{{24hR} \over {25m}}

Q52

If the nuclear radius of 27 Al is 3.6 fermi, the approximate nuclear radius of 64 Cu in fermi is

A 2.4

B 1.2

C 4.8

D 3.6

Correct Answer

Option C

Solution

Nuclear radius, R = R 0 A 1/3 where R 0 is a constant and A is the mass number $\therefore$

{{{R_{Al}}} \over {{R_{Cu}}}} = {{{{\left( {27} \right)}^{1/3}}} \over {{{\left( {64} \right)}^{1/3}}}}

=

{3 \over 4}

$\Rightarrow$ R Cu =

{4 \over 3} \times {R_{Al}}

=

{4 \over 3} \times 3.6

= 4.8 fermi

Q53

Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths

\lambda

1 :

\lambda

2 emitted in the two cases is

A

{7 \over 5}

B

{27 \over 20}

C

{27 \over 5}

D

{20 \over 7}

Correct Answer

Option D

Solution

In first case, n 1 = 3 and n 2 = 4

{1 \over {{\lambda _1}}} = R\left( {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right)

=

{{7R} \over 144}

In second case, n 1 = 2 and n 2 = 3

{1 \over {{\lambda _2}}} = R\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right)

=

{{5R} \over {36}}

$\therefore$

{{{\lambda _1}} \over {{\lambda _2}}}

=

{5 \over {36}} \times {{144} \over 7}

=

{20 \over 7}

Q54

An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is

A 2

B 3

C 4

D 5

Correct Answer

Option C

Solution

KE max = 10eV $\phi$ = 2.75 eV Total incident energy E = $\phi$ + KE max = 12.75 eV $\therefore$ Energy is released when electron jumps from the excited state n to the ground state.

$\therefore$ E 4 – E 1 = {– 0.85 – (–13.6) ev} = 12.75eV $\therefore$ value of n = 4

Q55

Two radioactive nuclei P and Q, in a given sample decay into a stable nucleus R. At time t = 0. number of P species are 4 N 0 and that of Q are N 0 . Half -life of P (for conversion to R) is 1 minute where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of R present in the sample would be

A 2 N 0

B 3 N 0

C

{{9{N_0}} \over 2}

D

{{5{N_0}} \over 2}

Correct Answer

Option C

Solution

Initially P = 4N 0 Q = N 0 Half life T P = 1 min.

T Q = 2 min.

Let after time t number of nuclei of P and Q are equal, that is N P = N Q $\Rightarrow$

4{N_0}{\left( {{1 \over 2}} \right)^{t/1}}

=

{N_0}{\left( {{1 \over 2}} \right)^{t/2}}

$\Rightarrow$

{2^{t/1}} = {4.2^{t/2}}

$\Rightarrow$ t = 4 min After 4 minutes, both P and Q have equal number of nuclei. $\therefore$ Number of nuclei of R =

\left( {4{N_0} - {{{N_0}} \over 4}} \right)

+

\left( {{N_0} - {{{N_0}} \over 4}} \right)

=

{{9{N_0}} \over 2}

Q56

Out of the following which one is not a possible energy for a photon to be emitted by hydrogen atom according to Bohr's atomic model?

A 0.65 eV

B 1.9 eV

C 11.1 eV

D 13.6 eV

Correct Answer

Option C

Solution

E n =

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

$\therefore$ E 1 = –13.6 eV E 2 = –3.4 eV E 3 = –1.5 eV E 4 = –0.85 eV $\therefore$ E 3 – E 2 = –1.5 – (–3.4) = 1.9 eV E 4 – E 3 = –0.85 – (–1.5) = 0.65 eV Obviously, difference of 11.1 eV is not possible.

Q57

Fusion reaction takes place at high temperature because

A nuclei break up at high temperature

B atoms get ionised at high temperature

C kinetic energy is high enough to overcome the coulomb repulsion between nuclei

D molecules break up at high temperature

Correct Answer

Option C

Solution

Extremely high temperature needed for fusion make kinetic energy large enough to overcome coulomb repulsion between nuclei.

Q58

A nucleus

{}_n^mX

emits one

\alpha

particle and two

\beta

particles. The resulting nucleus is

A

{}_{n - 4}^{m - 6}Z

B

{}_n^{m - 6}Z

C

{}_n^{m - 4}X

D

{}_{n - 2}^{m - 4}Y

Correct Answer

Option C

Solution

$\alpha$ -particle is 2 He 4 In $\beta$ emission, the neutron gets converted to proton and electron n X m

\overset{{1\alpha }}\longrightarrow

n - 2 Y m - 4

\overset{{2\beta }}\longrightarrow

n Z m - 4

Q59

A radioactive nucleus of mass M emits a photon of frequency

v

and the nucleus recoils. The recoil energy will be

A Mc 2

-

h

v

B h 2

\upsilon

2 /2Mc 2

C zero

D h

v

Correct Answer

Option B

Solution

Momentum Mu =

{E \over c} = {{hv} \over c}

Recoil energy :

{1 \over 2}M{u^2}

=

{1 \over 2}{{{M^2}{u^2}} \over M}

=

{1 \over {2M}}{\left( {{{hv} \over c}} \right)^2}

=

{{{h^2}{\nu ^2}} \over {2Mc}}

Q60

The power obtained in a reactor using U 235 disintegration is 1000 kW. The mass decay of U 235 per hour is

A 10 microgram

B 20 microgram

C 40 microgram

D 1 microgram

Correct Answer

Option C

Solution

From Einstein relation, E = mc 2 Rearranging, m =

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

We see that mass decay per second :

{{dm} \over {dt}}

=

{1 \over {{c^2}}}

×

{{dE} \over {dt}}

=

{{1000 \times {{10}^3}} \over {{{\left( {3 \times {{10}^8}} \right)}^2}}}

Now mass decay of U 235 per hour, =

{{dm} \over {dt}}

$\times$ 60 $\times$ 60 =

{{1000 \times {{10}^3}} \over {{{\left( {3 \times {{10}^8}} \right)}^2}}}

$\times$ 3600 = 4 × 10 –8 kg = 40 microgram