Atoms and Nuclei — Page 2 | JEE Physics

Q11

The number of spectral lines emitted by atomic hydrogen that is in the 4th energy level, is

A 3

B 6

C 1

D 0

Correct Answer

Option B

Solution

When an e transition from n $^{th}$ to ground state then

\sum {(n - 1)}

spectral lines are produced. Given, $n=4$ So,

\sum {(4 - 1) = \sum {3 = 3 + 2 + 1 = 6} }

Hence, 6 spectral lines are emitted by atomic Hydrogen that is in the $4^{th}$ energy level

Q12

The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as:

A

\dfrac{1}{n^4}

B

\dfrac{1}{n^2}

C

\dfrac{1}{n^3}

D

\dfrac{1}{n}

Correct Answer

Option C

Solution

Frequency of revolution $\propto \dfrac{1}{\mathrm{n}^3}$

Q13

For a nucleus of mass number A and radius R, the mass density of nucleus can be represented as

A

A^{\dfrac{2}{3}}

B Independent of A

C

A^3

D

A^{\dfrac{1}{3}}

Correct Answer

Option B

Solution

The radius $R$ of a nucleus is given by the empirical relation:

R = r_0 A^{\frac{1}{3}},

where $r_0$ is a constant and $A$ is the mass number.

The volume $V$ of the nucleus is approximately that of a sphere:

V = \frac{4}{3}\pi R^3.

Substituting the expression for $R$ into the volume formula:

V = \frac{4}{3}\pi \left(r_0 A^{\frac{1}{3}}\right)^3 = \frac{4}{3}\pi r_0^3 A.

The mass density $\rho$ is defined as the mass (which is proportional to $A$ ) divided by the volume:

\rho = \frac{A}{V} = \frac{A}{\frac{4}{3}\pi r_0^3 A} = \frac{1}{\frac{4}{3}\pi r_0^3}.

Notice that $A$ cancels out in the numerator and denominator, so the density is:

\rho = \text{constant},

meaning it is independent of the mass number $A$ . Thus, the correct option is: Option B: Independent of A.

Q14

In a radioactive decay chain, the initial nucleus is

{}_{90}^{232}

Th. At the end there are 6

\alpha

-particles and 4

\beta

-particles which are emitted. If the end nucleus is

{}_Z^A

X, A and Z are given by :

A A = 208; Z = 80

B A = 208; Z = 82

C A = 200; Z = 81

D A = 202; Z = 80

Correct Answer

Option B

Solution

{}_{90}^{232}

Th

\overset{\,}\longrightarrow

{}_{78}^{208}

Y +

{}_2^4

He

{}_{78}^{208}

Y

\overset{\,}\longrightarrow

{}_{82}^{208}

X + 4 $\beta$

Q15

If an electron is moving in the nth orbit of the hydrogen atom, then its velocity (vn) for the nth orbit is given as :

A

{v_n} \propto {1 \over n}

B vn

\propto

n2

C vn

\propto

n

D

{v_n} \propto {1 \over {{n^2}}}

Correct Answer

Option A

Solution

We know velocity of electron in nth shell of hydrogen atom is given by

v = {{2\pi kZ{e^2}} \over {nh}}

$\therefore$

v \propto {1 \over n}

Q16

If the binding energy of the electron in a hydrogen atom is

13.6eV,

the energy required to remove the electron from the first excited state of

L{i^{ + + }}

is

A

30.6

eV

B

13.6

eV

C

3.4

eV

D

122.4

eV

Correct Answer

Option A

Solution

{E_n} = - {{13.6} \over {{n^2}}}{Z^2}eV/

atom For lithium ion

Z=3;

n=2

(for first excited state)

{E_n} = - {{13.6} \over {{2^2}}} \times {3^2} = - 30.6eV

Q17

A radioactive sample at any instant has its disintegration rate

5000

disintegrations per minute. After

5

minutes, the rate is

1250

disintegrations per minute. Then, the decay constant (per minute) is

A

0.4

ln2

B

0.2

ln2

C

0.1

ln2

D

0.8

ln2

Correct Answer

Option A

Solution

\lambda = {1 \over t}{\log _e}{{{A_0}} \over A}

= {1 \over 5}{\log _e}{{5000} \over {1250}}

= 0.2{\log _e}4

= 0.4{\log _e}2

Q18

If

{N_0}

is the original mass of the substance of half-life period

{t_{1/2}} = 5

years, then the amount of substance left after

15

years is

A

{N_0}/8

B

{N_0}/16

C

{N_0}/2

D

{N_0}/4

Correct Answer

Option A

Solution

After every half-life, the mass of the substance reduces to half its initial value.

{N_0}\mathop \to \limits^{5\,years} \,\,{{{N_0}} \over 2}\mathop \to \limits^{5\,years} {{{N_0}} \over {{2^2}}}\mathop \to \limits^{5\,years} {{{N_0}} \over 8}

Q19

Which of the following radiations has the least wavelength ?

A

\gamma

- rays

B

\beta

- rays

C

\alpha

- rays

D

X

- rays

Correct Answer

Option A

Solution

The electromagnetic spectrum is as follows $\therefore$ $\gamma$ -rays has least wavelength.

Q20

If the binding energy per nucleon in

{}_3^7Li

and

{}_2^4He

nuclei are

5.60

MeV

and

7.06

MeV

respectively, then in the reaction

p + {}_3^7Li \to 2\,{}_2^4He

$ energy of proton must be

A

28.24

MeV

B

17.28

MeV

C

1.46

MeV

D

39.2

MeV

Correct Answer

Option B

Solution

Let

E

be the energy of proton, then

E + 7 \times 5.6 = 2 \times \left[ {4 \times 7.06} \right]

\Rightarrow E = 56.48 - 39.2 = 17.28MeV