Atoms and Nuclei — Page 19 | JEE Physics

Q181

Considering Bohr's atomic model for hydrogen atom : (A) the energy of H atom in ground state is same as energy of

\mathrm{He}^{+}

ion in its first excited state. (B) the energy of H atom in ground state is same as that for

\mathrm{Li}^{++}

ion in its second excited state. (C) the energy of H atom in its ground state is same as that of

\mathrm{He}^{+}

ion for its ground state. (D) the energy of

\mathrm{He}^{+}

ion in its first excited state is same as that for

\mathrm{Li}^{++}

ion in its ground state. Choose the correct answer from the options given below :

A (A), (B) only

B (A), (D) only

C (A), (C) only

D (B), (D) only

Correct Answer

Option A

Solution

In Bohr's atomic model, the energy of an electron in an atom is given by the formula: $E \propto \dfrac{Z^2}{n^2}$ where $Z$ is the atomic number and $n$ is the principal quantum number of the electron's orbit.

Let’s consider the implications: For hydrogen ( $\mathrm{H}$ ), $Z = 1$ .

For helium ion ( $\mathrm{He}^{+}$ ), $Z = 2$ .

For lithium ion ( $\mathrm{Li}^{++}$ ), $Z = 3$ .

Now, identifying the relevant states: Ground State: This corresponds to $n = 1$ .

First Excited State: This corresponds to $n = 2$ .

Second Excited State: This corresponds to $n = 3$ .

Let's evaluate the energy comparisons: Hydrogen Atom in Ground State: $E \propto \dfrac{1^2}{1^2} = 1$ .

Helium Ion ( $\mathrm{He}^{+}$ ) in the First Excited State: $E \propto \dfrac{2^2}{2^2} = 1$ This matches the energy of hydrogen in its ground state.

Lithium Ion ( $\mathrm{Li}^{++}$ ) in the Second Excited State: $E \propto \dfrac{3^2}{3^2} = 1$ This also matches the energy of hydrogen in its ground state.

From this analysis, we can conclude: The energy of a hydrogen atom in its ground state is equal to the energy of a $\mathrm{He}^{+}$ ion in its first excited state.

The energy of a hydrogen atom in its ground state also equals the energy of a $\mathrm{Li}^{++}$ ion in its second excited state.

Hence, statements (A) and (B) are correct.

Q182

In a hydrogen like ion, the energy difference between the

2^{\text{nd }}

excitation energy state and ground is 108.8 eV . The atomic number of the ion is:

A 1

B 4

C 3

D 2

Correct Answer

Option C

Solution

To determine the atomic number (Z) of a hydrogen-like ion, given that the energy difference between its second excitation state and the ground state is 108.8 eV, we use the following formula for the energy difference between two energy levels in a hydrogen-like atom: $\Delta E = 13.6 \, \text{eV} \times Z^2 \left[\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right]$ Here, for the second excitation, $n_2 = 3$ (since excitation means moving to the second higher energy level beyond the ground state, which means $n_1 = 1$ for the ground state to $n_2 = 3$ ).

Plug these into the energy difference formula: $\Delta E = 13.6 \, Z^2 \left[\dfrac{1}{1^2} - \dfrac{1}{3^2}\right]$ Simplifying the term in brackets: $\Delta E = 13.6 \, Z^2 \left[1 - \dfrac{1}{9}\right] = 13.6 \, Z^2 \times \dfrac{8}{9}$ Given that $\Delta E = 108.8 \, \text{eV}$ , set the equations equal: $13.6 \, \dfrac{8Z^2}{9} = 108.8$ Solving for $Z$ , first simplify: $13.6 \times \dfrac{8}{9} \times Z^2 = 108.8$ $\dfrac{108.8 \times 9}{13.6 \times 8} = Z^2$ $Z^2 = 9$ Thus, the atomic number $Z$ is: $Z = 3$ The atomic number of the ion is 3.

Q183

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is

A

5: 27

B

27: 5

C

3: 4

D

5: 36

Correct Answer

Option A

Solution

Lyman

\begin{aligned} & \frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]=\frac{3 \mathrm{R}}{4} \\ & \lambda_1=\frac{4}{3 \mathrm{R}}\quad\text{..... (1)} \end{aligned}

and Balmer

\begin{aligned} &\begin{aligned} & \frac{1}{\lambda_2}=\mathrm{R}\left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 \mathrm{R}}{36} \\ & \lambda_2=\frac{36}{5 \mathrm{R}} \end{aligned}\\ &\text{ Then, } \frac{\lambda_1}{\lambda_2}=\frac{5}{27} \end{aligned}

Q184

Energy released when two deuterons

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

fuse to form a helium nucleus

\left({ }_2 \mathrm{He}^4\right)

is : (Given : Binding energy per nucleon of

{ }_1 \mathrm{H}^2=1.1 \mathrm{MeV}

and binding energy per nucleon of

{ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}

)

A 26.8 MeV

B 8.1 MeV

C 23.6 MeV

D 5.9 MeV

Correct Answer

Option C

Solution

To calculate the energy released when two deuterons ( ${}_1 \mathrm{H}^2$ ) fuse to form a helium nucleus ( ${}_2 \mathrm{He}^4$ ), consider the following: The reaction is represented as:

{}_1{H^2} + {}_1{H^2} \to {}_2{He^4}

Given the binding energy per nucleon: For ${}_1 \mathrm{H}^2$ : 1.1 MeV For ${}_2 \mathrm{He}^4$ : 7.0 MeV Calculating Binding Energy Binding Energy for Reactants: Each deuteron ( ${}_1 \mathrm{H}^2$ ) has a binding energy of $1.1 \, \text{MeV}$ per nucleon.

Since there are two nucleons in a deuteron, the total binding energy for one deuteron is $1.1 \times 2 = 2.2 \, \text{MeV}$ .

Therefore, for two deuterons: $2.2 \times 2 = 4.4 \, \text{MeV}$ .

Binding Energy for Product: For ${}_2 \mathrm{He}^4$ : Each of the four nucleons has a binding energy of $7.0 \, \text{MeV}$ .

Total binding energy for the helium nucleus: $7.0 \times 4 = 28.0 \, \text{MeV}$ .

Energy Released (Q) The energy released, $\mathrm{Q}$ , is the difference between the binding energy of the products and the reactants: $\mathrm{Q} = \mathrm{BE}_{\text{product}} - \mathrm{BE}_{\text{reactants}} = 28.0 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV}$ Thus, the energy released when two deuterons fuse to form a helium nucleus is 23.6 MeV.

Q185

Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of

\mathrm{Li}^{++}

ion in its ground state is given by

\dfrac{1}{X} a_0

, where

X=

__________ (Where

\mathrm{a}_0

is the first Bohr's radius.)

A 2

B 9

C 1

D 3

Correct Answer

Option D

Solution

The radius of a hydrogen-like ion according to Bohr's atomic model is given by the formula: $r = r_0 \dfrac{n^2}{Z}$ Where: $r_0$ is the first Bohr radius ( $a_0$ ) $n$ represents the principal quantum number $Z$ is the atomic number of the ion For the ion $\text{Li}^{++}$ , the atomic number $Z = 3$ and we are considering the ground state, so $n = 1$ .

Plugging these values into the formula, we get: $r = r_0 \dfrac{1^2}{3} = \dfrac{r_0}{3}$ This shows that the radius of $\text{Li}^{++}$ in its ground state is $\dfrac{1}{3} a_0$ , meaning $X = 3$ .

Q186

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The density of the copper

(^ {64}_{29} \text{Cu})

nucleus is greater than that of the carbon

(^ {12}_{6} \text{C})

nucleus. Reason (R): The nucleus of mass number A has a radius proportional to

A^{1/3}

. In the light of the above statements, choose the most appropriate answer from the options given below:

A (A) is correct but (R) is not correct

B Both (A) and (R) are correct but (R) is not the correct explanation of (A)

C (A) is not correct but (R) is correct

D Both (A) and (R) are correct and (R) is the correct explanation of (A)

Correct Answer

Option C

Solution

To evaluate the assertion and reason provided, we need to consider the formula for the density of a nucleus.

The density $\rho$ can be expressed as follows: $\rho = \dfrac{M}{V} = \dfrac{m_n \times A}{\dfrac{4}{3} \pi R^3}$ Here, $m_n$ represents the nucleon mass, $A$ is the mass number, and $R$ is the radius of the nucleus.

According to the formula for the radius: $R = R_0 A^{1/3}$ This indicates that the radius $R$ is proportional to $A^{1/3}$ .

By substituting the expression for $R$ in terms of $A$ back into the formula for density, we get: $\rho = \dfrac{m_n \times A}{\dfrac{4}{3} \pi (A^{1/3} R_0)^3}$ Simplifying the equation shows that the $A$ terms cancel out: $\rho = \dfrac{m_n}{\dfrac{4}{3} \pi R_0^3}$ This reveals that the nuclear density $\rho$ is approximately constant and independent of the mass number $A$ .

Therefore, the densities of copper $(^{64}_{29} \text{Cu})$ and carbon $(^{12}_{6} \text{C})$ nuclei are effectively the same because the density formula results in a similar value regardless of the specific mass number.

Q187

Given below are two statements : Statement (I) : The dimensions of Planck's constant and angular momentum are same. Statement (II) : In Bohr's model electron revolve around the nucleus only in those orbits for which angular momentum is integral multiple of Planck's constant. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are correct

B Statement I is correct but Statement II is incorrect

C Both Statement I and Statement II are incorrect

D Statement I is incorrect but Statement II is correct

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & \mathrm{E}=\mathrm{hf} \\ & \mathrm{ML}^2 \mathrm{~T}^{-2}=[\mathrm{h}] \times\left[\mathrm{T}^{-1}\right] \\ & {[\mathrm{h}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]} \\ & \mathrm{L}=[\mathrm{MVR}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\ & \mathrm{L}=\frac{\mathrm{nh}}{2 \pi} \end{aligned}\\ &\mathrm{L} \text{ is integral multiple of } \frac{\mathrm{h}}{2 \pi} \end{aligned}

Q188

Considering the Bohr model of hydrogen like atoms, the ratio of the radius of

5^{\text{th }}

orbit of the electron in

\mathrm{Li}^{2+}

and

\mathrm{He}^{+}

is

A

\dfrac{3}{2}

B

\dfrac{2}{3}

C

\dfrac{4}{9}

D

\dfrac{9}{4}

Correct Answer

Option B

Solution

In the Bohr model for a hydrogen-like atom, the radius of the $n^{\text{th}}$ orbit is given by

r_n = \frac{n^2 a_0}{Z}

where: $n$ is the principal quantum number, $a_0$ is the Bohr radius, and $Z$ is the atomic number (the effective nuclear charge).

For the $5^{\text{th}}$ orbit ( $n = 5$ ): For $\mathrm{Li}^{2+}$ (where $Z = 3$ ):

r_5(\mathrm{Li}^{2+}) = \frac{5^2 a_0}{3} = \frac{25a_0}{3}

For $\mathrm{He}^{+}$ (where $Z = 2$ ):

r_5(\mathrm{He}^{+}) = \frac{5^2 a_0}{2} = \frac{25a_0}{2}

Now, the ratio of the radius of the 5th orbit of $\mathrm{Li}^{2+}$ to that of $\mathrm{He}^{+}$ is:

\frac{r_5(\mathrm{Li}^{2+})}{r_5(\mathrm{He}^{+})} = \frac{\frac{25a_0}{3}}{\frac{25a_0}{2}} = \frac{25a_0}{3} \times \frac{2}{25a_0} = \frac{2}{3}

Thus, the ratio is $\dfrac{2}{3}$ . The correct answer is Option B.

Q189

A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is :

A 2

B 3

C 4

D 5

Correct Answer

Option C

Solution

Energy released when hydrogen atom makes transition from n = 2 to n = 1 is, E1 = 13.6 $\times$

\left( {{1 \over {{1^2}}} - {1 \over {{z^2}}}} \right)

=

{3 \over 4} \times 13.6\,\,\,

eV Energy required to remove a electron from nth excited state of doubly ionized lithium, E2 =

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

=

{{13.6 \times {3^2}} \over {{n^2}}}

eV This energy is provided by the photon when it strike with the lithium atom. $\therefore$ E1 $\ge$ E2 $\Rightarrow$

{3 \over 4} \times 13.6 \ge {{13.6 \times 9} \over {{n^2}}}

$\Rightarrow$ n2 $\ge$ 3 $\times$ 4 $\Rightarrow$ n $\ge$

\sqrt {12}

$\Rightarrow$ n $\ge$ 3.5 $\therefore$ Least possible excited state = 4

Q190

If the series limit frequency of the Lyman series is

{\nu _L}

, then the series limit frequency of the Pfund series is:

A

{\nu _L}/25

B

25{\nu _L}

C

16{\nu _L}

D

{\nu _L}/16

Correct Answer

Option A

Solution

Note : (1) In Lyman Series, transition happens in n = 1 state from n = 2, 3, . . . . .

$\propto$ (2) In Balmer Series, transition happens in n = 2 state from n = 3, 4, . . . . .

$\propto$ (3) In Paschen Series, transition happens in n = 3 state from n = 4, 5, . . . . .

$\propto$ (4) In Bracktt Series, transition happens in n = 4 state from n = 5, 6 . . . . . .

$\propto$ (5) In Pfund Series, transition happens in n = 5 state from n = 6, 7, . . . .

$\propto$ We know,

{1 \over \lambda }

= RZ2

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

Series limit means transition happens from n = $\propto$ to n = 1, for Lyman Series. In series limit for Lyman series,

{1 \over {{\lambda _L}}}

= RZ2

\left( {{1 \over {{1^2}}} - {1 \over \propto }} \right)

$\Rightarrow$

\,\,\,

{1 \over {{\lambda _L}}}

= RZ2 We know, E =

{{hc} \over \lambda }

= h $\gamma$ $\Rightarrow$

\,\,\,

$\gamma$ =

{c \over \lambda }

So, frequency in Lyman Series, $\gamma$ L =

{c \over {{\lambda _L}}}

= c $\times$ RZ2 In Pfund series, n2 = $\propto$ and n1 = 5

\therefore\,\,\,

{1 \over {{\lambda _P}}}

= RZ2

\left( {{1 \over {{5^2}}} - {1 \over {{ \propto ^2}}}} \right)

$\Rightarrow$

\,\,\,

{1 \over {{\lambda _P}}}

=

{{R{Z^2}} \over {25}}

\therefore\,\,\,

{\gamma _P}

=

{c \over {{\lambda _P}}}

= c $\times$

{{R{Z^2}} \over {25}}

\therefore\,\,\,

$\gamma$ P =

{{cRZ{}^2} \over {25}}

=

{{{\gamma _L}} \over {25}}

[as $\gamma$ L = cRZ2]