Atoms

In Bohr's model of the hydrogen atom, the kinetic energy (\(K\)) and potential energy (\(U\)) of an electron in an orbit are related to its distance from the nucleus. When an electron transitions from a higher energy level (\(n=2\)) to a lower energy level (\(n=1\)), it loses energy. The kinetic energy of the electron is given by the formula \(K = \frac{1}{2}mv^2\), and the potential energy is given by \(U = -\frac{ke^2}{r}\), where \(r\) is the radius of the orbit. As the electron moves to a lower orbit (\(n=1\)), the radius decreases, leading to an increase in kinetic energy (since the electron moves faster in lower orbits) and a decrease in potential energy (since it is closer to the nucleus). Therefore, \(K\) increases and \(U\) decreases.
Identify the initial and final orbits: initial orbit \(n=2\), final orbit \(n=1\).

Understand that as the electron moves to a lower orbit, it loses energy.

Determine that kinetic energy increases because the electron moves faster in the lower orbit.

Determine that potential energy decreases because the electron is closer to the nucleus

A. \(K\) increases and \(U\) decreases

$$\begin{gathered} \lambda =\frac{h}{mv} \\ mvr=\frac{nh}{2\pi },\lambda =\frac{2\pi rh}{nh} \\ \lambda \propto \frac{r}{n} \\ \frac{{\lambda }_1}{{\lambda }_4}=\frac{r_1{n}_4}{n_1{r}_4}=\frac{5.3\times {10}^{-11}\times 4}{1\times 84.8\times {10}^{-11}} \\ =\frac{1}{4} \end{gathered}$$

In the alpha particle scattering experiment, the impact parameter is the perpendicular distance between the path of an alpha particle and the center of the nucleus. The scattering angle is the angle at which the alpha particle is deflected. The correct statements are as follows:

(A) For a scattering angle \(\theta \approx 0\), the impact parameter is large because the alpha particle passes far from the nucleus and experiences a small deflection.

(B) For a scattering angle \(\theta \approx \pi\), the impact parameter is small because the alpha particle passes very close to the nucleus and experiences a large deflection.

(C) The number of alpha particles undergoing head-on collision is small because head-on collisions are rare events.

(D) The experiment provides an estimate of the upper limit to the size of the target nucleus because the closest approach of the alpha particles gives information about the size of the nucleus.

\begin{equation}
\begin{aligned}
& \mathrm{P}_1=\mathrm{P}=\frac{\mathrm{N}_1 \mathrm{hc}}{\lambda_1} \quad \mathrm{P}=\frac{\mathrm{Nhc}}{\lambda} \\
& \mathrm{P}_2=\frac{\mathrm{P}}{2}=\frac{\mathrm{N}_2 \mathrm{hc}}{\lambda_2} \\
& \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{N}_1}{\lambda_1} \cdot \frac{\lambda_2}{\mathrm{~N}_2} \\
& \mathrm{~N}_2=\frac{\mathrm{N}_1 \lambda_2}{\lambda_1 2}=\frac{2 \times 10^{15} \times 300}{600 \times 2} \\
& \mathrm{n}_2=5 \times 10^{14} \text { per second }
\end{aligned}
\end{equation}

For the first excited state of a hydrogen atom ( n=2), the kinetic energy is 3.4 eV and the potential energy is

$-6.8$eV. This is calculated using the formula for energy levels ($E_n=-13.6/n^2$Sv6Kpe[] eV) and the relationship that potential energy ($U$Sv6Kpe[]) is equal to $-2$Sv6Kpe[] times the kinetic energy ($K$Sv6Kpe[]), with total energy ($E$Sv6Kpe[]) being the sum of both ($E=K+U$Sv6Kpe[]), and kinetic energy being the negative of the total energy ($K=-E$)

In hydrogen-like atoms, energy of the \( n^\text{th} \) level is:

\[
E_n = -13.6 \cdot \frac{Z^2}{n^2}
\]

Second excitation means third level \( (n = 3) \), and ground state is \( n = 1 \).

So,

\[
\Delta E = E_1 - E_3 = 13.6 Z^2 \left(1 - \frac{1}{9}\right)
= 13.6 Z^2 \cdot \frac{8}{9}
\]

Given: \( \Delta E = 108.8 \, \text{eV} \)

\[
\Rightarrow 13.6 Z^2 \cdot \frac{8}{9} = 108.8
\Rightarrow Z^2 = \frac{108.8 \cdot 9}{13.6 \cdot 8} = 9
\Rightarrow Z = 3
\]

In the alpha particle scattering experiment, the impact parameter is the perpendicular distance between the path of an alpha particle and the center of the nucleus. The scattering angle is the angle at which the alpha particle is deflected. The correct statements are as follows:

(A) For a scattering angle \(\theta \approx 0\), the impact parameter is large because the alpha particle passes far from the nucleus and experiences a small deflection.

(B) For a scattering angle \(\theta \approx \pi\), the impact parameter is small because the alpha particle passes very close to the nucleus and experiences a large deflection.

(C) The number of alpha particles undergoing head-on collision is small because head-on collisions are rare events.

(D) The experiment provides an estimate of the upper limit to the size of the target nucleus because the closest approach of the alpha particles gives information about the size of the nucleus.

• The electric potential energy ($U$) between the alpha particle (charge$\mathrm{+2}e$) and the gold nucleus (charge$+Ze$) at a distance $d$is given by $U=k\frac{\left(2e\right)\left(Ze\right)}{d}=\frac{2kZ{e}^2}{d}$, where $k$is Coulomb's constant.

• Equating energies: By setting $K=U$, we get $K=\frac{2kZ{e}^2}{d}$ $d=\frac{2kZ{e}^2}{K}$ the equation shows that $d$is directly proportional to $\frac{1}{K}$

For hydrogen atom, the Balmer series corresponds to transitions ending at $n_2 =2$
Thus, the general transition is $n_i\to 2, where n_i =3,4,5...$
The first Balmer line $\left(H_{\alpha }\right)$ : $n_i =3\to 2$
The second Balmer line ($H_{\beta }$) : $n_i =4 \to 2$
Hence, the transition giving the second Balmer line is $n_i =4 \to n_f =2$

In Bohr's model of the hydrogen atom, the kinetic energy (\(K\)) and potential energy (\(U\)) of an electron in an orbit are related to its distance from the nucleus. When an electron transitions from a higher energy level (\(n=2\)) to a lower energy level (\(n=1\)), it loses energy. The kinetic energy of the electron is given by the formula \(K = \frac{1}{2}mv^2\), and the potential energy is given by \(U = -\frac{ke^2}{r}\), where \(r\) is the radius of the orbit. As the electron moves to a lower orbit (\(n=1\)), the radius decreases, leading to an increase in kinetic energy (since the electron moves faster in lower orbits) and a decrease in potential energy (since it is closer to the nucleus). Therefore, \(K\) increases and \(U\) decreases.
Identify the initial and final orbits: initial orbit \(n=2\), final orbit \(n=1\).

Understand that as the electron moves to a lower orbit, it loses energy.

Determine that kinetic energy increases because the electron moves faster in the lower orbit.

Determine that potential energy decreases because the electron is closer to the nucleus

A. \(K\) increases and \(U\) decreases

$$\begin{gathered} r_n= \frac{n^{2}h{\epsilon }_0}{\pi m{e}^2 } \\ {r}_n \alpha n^2 \end{gathered}$$

For the first excited state of a hydrogen atom ( n=2), the kinetic energy is 3.4 eV and the potential energy is

$-6.8$eV. This is calculated using the formula for energy levels ($E_n=-13.6/n^2$Sv6Kpe[] eV) and the relationship that potential energy ($U$Sv6Kpe[]) is equal to $-2$Sv6Kpe[] times the kinetic energy ($K$Sv6Kpe[]), with total energy ($E$Sv6Kpe[]) being the sum of both ($E=K+U$Sv6Kpe[]), and kinetic energy being the negative of the total energy ($K=-E$)

As the wavelength of spectral lines in the Balmer Series of H-atoms drops, they appear close together and have a lower intensity due to transition probability.

In Bohr's model, radius of \( n^\text{th} \) orbit in hydrogen-like atoms is given by:

\[
r_n \propto \frac{n^2}{Z}
\]

So for 5th orbit:

\[
r_5 \propto \frac{25}{Z}
\]

Now,
For Li\(^{2+}\), \( Z = 3 \Rightarrow r \propto \frac{25}{3} \)
For He\(^{+}\), \( Z = 2 \Rightarrow r \propto \frac{25}{2} \)

Taking ratio:

\[
\frac{r_{Li^{2+}}}{r_{He^+}} = \frac{25/3}{25/2} = \frac{2}{3}
\]

In Bohr's model of the hydrogen atom, the kinetic energy (\(K\)) and potential energy (\(U\)) of an electron in an orbit are related to its distance from the nucleus. When an electron transitions from a higher energy level (\(n=2\)) to a lower energy level (\(n=1\)), it loses energy. The kinetic energy of the electron is given by the formula \(K = \frac{1}{2}mv^2\), and the potential energy is given by \(U = -\frac{ke^2}{r}\), where \(r\) is the radius of the orbit. As the electron moves to a lower orbit (\(n=1\)), the radius decreases, leading to an increase in kinetic energy (since the electron moves faster in lower orbits) and a decrease in potential energy (since it is closer to the nucleus). Therefore, \(K\) increases and \(U\) decreases.
Identify the initial and final orbits: initial orbit \(n=2\), final orbit \(n=1\).

Understand that as the electron moves to a lower orbit, it loses energy.

Determine that kinetic energy increases because the electron moves faster in the lower orbit.

Determine that potential energy decreases because the electron is closer to the nucleus

A. \(K\) increases and \(U\) decreases

As the wavelength of spectral lines in the Balmer Series of H-atoms drops, they appear close together and have a lower intensity due to transition probability.

Moseley's law states that $\sqrt{f}=k\left(z-b\right)$, where k is a constant and b is the screening constant. This implies that the square root of frequency is directly proportional to Z, resulting in a straight-line graph. However, f itself does not follow a linear relationship with Z. Moseley's law is primarily valid for heavy elements, and deviations occur for light atoms due to weaker screening effects and other relativistic considerations. Hence, both statements are correct.

No. of spectral line $=\frac{n(n-1)}{2}=\frac{4\times 3}{2}=6$

$$\begin{gathered} r_n= \frac{n^{2}h{\epsilon }_0}{\pi m{e}^2 } \\ {r}_n \alpha n^2 \end{gathered}$$

First member of Lyman series: \( n_2 = 2 \to n_1 = 1 \)
Second member: \( n_2 = 3 \to n_1 = 1 \)

From Rydberg formula:

\[
\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)
\]

For first member:
\[
\frac{1}{\lambda_1} = R \left(1 - \frac{1}{4} \right) = R \cdot \frac{3}{4}
\]

For second member:
\[
\frac{1}{\lambda_2} = R \left(1 - \frac{1}{9} \right) = R \cdot \frac{8}{9}
\]

Taking ratio:
\[
\frac{\lambda_2}{\lambda_1} = \frac{\frac{1}{R \cdot \frac{8}{9}}}{\frac{1}{R \cdot \frac{3}{4}}} = \frac{3}{4} \cdot \frac{9}{8} = \frac{27}{32}
\]

\[
\lambda_2 = \frac{27}{32} \lambda
\]

No. of spectral line $=\frac{n(n-1)}{2}=\frac{4\times 3}{2}=6$

1. Bohr’s Quantization Rule: The angular momentum of the electron is quantized:

   $$m v r = n \frac{h}{2\pi}$$​

   where:

   • $m$ is the mass of the electron,
   • $v$ is its velocity,
   • $r$ is the radius of the orbit,
   • $n$ is the quantum number ($n = 2$ for the 1st excited state),
   • $h$ is Planck's constant.

2. Magnetic Force Provides Centripetal Force: The force on a moving charge in a magnetic field is:

   $$F_{\text{magnetic}} = e v B$$

   This provides the centripetal force:

   $$F_{\text{magnetic}} = \frac{m v^2}{r}$$​

From the magnetic force equation:

Simplify to find $v$:

Substitute $v = \frac{e B r}{m}$​ into Bohr’s quantization rule:

Simplify:

Rearrange to solve for $r^2$:

Thus:

For the 1st excited state, $n = 2$:

(C) $\sqrt{\frac{h}{\pi Be}}$

for second line of Balmer series, ${\mathrm{n}}_{\mathrm{f}} = 2, {\mathrm{n}}_{\mathrm{i}} = 4$

In the alpha particle scattering experiment, the impact parameter is the perpendicular distance between the path of an alpha particle and the center of the nucleus. The scattering angle is the angle at which the alpha particle is deflected. The correct statements are as follows:

(A) For a scattering angle \(\theta \approx 0\), the impact parameter is large because the alpha particle passes far from the nucleus and experiences a small deflection.

(B) For a scattering angle \(\theta \approx \pi\), the impact parameter is small because the alpha particle passes very close to the nucleus and experiences a large deflection.

(C) The number of alpha particles undergoing head-on collision is small because head-on collisions are rare events.

(D) The experiment provides an estimate of the upper limit to the size of the target nucleus because the closest approach of the alpha particles gives information about the size of the nucleus.

From Bohr's model, energy of hydrogen-like species is:

\[
E_n = -13.6 \cdot \frac{Z^2}{n^2} \, \text{eV}
\]

**(A)**
H atom ground state: \( Z = 1, n = 1 \Rightarrow E = -13.6 \, \text{eV} \)
He\(^+\) first excited state: \( Z = 2, n = 2 \Rightarrow E = -13.6 \cdot \frac{4}{4} = -13.6 \, \text{eV} \) ✅

**(B)**
H ground: \( -13.6 \, \text{eV} \)
Li\(^{++}\) second excited state: \( Z = 3, n = 3 \Rightarrow E = -13.6 \cdot \frac{9}{9} = -13.6 \, \text{eV} \) ✅

**(C)**
H ground: \( -13.6 \, \text{eV} \)
He\(^+\) ground: \( Z = 2, n = 1 \Rightarrow E = -13.6 \cdot 4 = -54.4 \, \text{eV} \) ❌

**(D)**
He\(^+\) first excited: \( Z = 2, n = 2 \Rightarrow E = -13.6 \cdot \frac{4}{4} = -13.6 \, \text{eV} \)
Li\(^{++}\) ground: \( Z = 3, n = 1 \Rightarrow E = -13.6 \cdot 9 = -122.4 \, \text{eV} \) ❌

Correct statements: **(A), (B)**

$$\begin{gathered} r_n= \frac{n^{2}h{\epsilon }_0}{\pi m{e}^2 } \\ {r}_n \alpha n^2 \end{gathered}$$

• The second excited state corresponds to the third energy level, where the principal quantum number $n=3$ $E_3=\frac{-13.6}{3^2}=\frac{-13.6}{9}\approx -1.51$eV.
• The potential energy ($PE$) is related to the total energy ($E$) by the formula $PE=2E$ Therefore, the potential energy in the second excited state is $P{E}_3=2\times E_3=2\times \left(-1.51\text{eV}\right)=-3.02$eV.

• The electric potential energy ($U$) between the alpha particle (charge$\mathrm{+2}e$) and the gold nucleus (charge$+Ze$) at a distance $d$is given by $U=k\frac{\left(2e\right)\left(Ze\right)}{d}=\frac{2kZ{e}^2}{d}$, where $k$is Coulomb's constant.

• Equating energies: By setting $K=U$, we get $K=\frac{2kZ{e}^2}{d}$ $d=\frac{2kZ{e}^2}{K}$ the equation shows that $d$is directly proportional to $\frac{1}{K}$

For the first excited state of a hydrogen atom ( n=2), the kinetic energy is 3.4 eV and the potential energy is

$-6.8$eV. This is calculated using the formula for energy levels ($E_n=-13.6/n^2$Sv6Kpe[] eV) and the relationship that potential energy ($U$Sv6Kpe[]) is equal to $-2$Sv6Kpe[] times the kinetic energy ($K$Sv6Kpe[]), with total energy ($E$Sv6Kpe[]) being the sum of both ($E=K+U$Sv6Kpe[]), and kinetic energy being the negative of the total energy ($K=-E$)

For the first excited state of a hydrogen atom ( n=2), the kinetic energy is 3.4 eV and the potential energy is

$-6.8$eV. This is calculated using the formula for energy levels ($E_n=-13.6/n^2$Sv6Kpe[] eV) and the relationship that potential energy ($U$Sv6Kpe[]) is equal to $-2$Sv6Kpe[] times the kinetic energy ($K$Sv6Kpe[]), with total energy ($E$Sv6Kpe[]) being the sum of both ($E=K+U$Sv6Kpe[]), and kinetic energy being the negative of the total energy ($K=-E$)

**Statement I**:
Planck’s constant \( h \) has dimensions of angular momentum:

\[
[h] = [M L^2 T^{-1}] = \text{same as angular momentum}
\]

✅ So, Statement I is **correct**.

**Statement II**:
In Bohr’s model, angular momentum is quantized as:

\[
mvr = n h / (2\pi) = n \hbar
\]

Which means electron revolves only in those orbits where angular momentum is integral multiple of Planck's constant \( \hbar \).
✅ So, Statement II is **also correct**.

**Conclusion**: Both statements are correct.

• The electric potential energy ($U$) between the alpha particle (charge$\mathrm{+2}e$) and the gold nucleus (charge$+Ze$) at a distance $d$is given by $U=k\frac{\left(2e\right)\left(Ze\right)}{d}=\frac{2kZ{e}^2}{d}$, where $k$is Coulomb's constant.

• Equating energies: By setting $K=U$, we get $K=\frac{2kZ{e}^2}{d}$ $d=\frac{2kZ{e}^2}{K}$ the equation shows that $d$is directly proportional to $\frac{1}{K}$

As the wavelength of spectral lines in the Balmer Series of H-atoms drops, they appear close together and have a lower intensity due to transition probability.

for second line of Balmer series, ${\mathrm{n}}_{\mathrm{f}} = 2, {\mathrm{n}}_{\mathrm{i}} = 4$

The radius of the $n$nth orbit in the Bohr model is given by:

$$r_n = n^2 r_1$$​

where:

• $r_n$​ is the radius of the $n$nth orbit.
• $r_1$​ is the radius of the first orbit.
• $n$ is the principal quantum number.

• The radius of the first Bohr orbit is $a$.
• We need to find the radius of the second Bohr orbit ($n = 2$).

Using the formula:

$$r_2 = 2^2 r_1 = 4 r_1$$

​ $$r_2 = 4a$$

$$\boxed{B} \quad 4a$$

Moseley's law states that $\sqrt{f}=k\left(z-b\right)$, where k is a constant and b is the screening constant. This implies that the square root of frequency is directly proportional to Z, resulting in a straight-line graph. However, f itself does not follow a linear relationship with Z. Moseley's law is primarily valid for heavy elements, and deviations occur for light atoms due to weaker screening effects and other relativistic considerations. Hence, both statements are correct.

The wavelength of light emitted during an electronic transition in the hydrogen atom is determined using the Rydberg formula:

$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right),$$

where:

• $R = 1.097 \times 10^7 \, \text{m}^{-1}$ (Rydberg constant),
• $n_1$​ and $n_2$ are the principal quantum numbers ($n_2 > n_1$​).

• Lyman Series: Transitions to $n_1 = 1$ (UV region).
• Balmer Series: Transitions to $n_1 = 2$ (visible region).
• Paschen Series: Transitions to $n_1 = 3$ (infrared region).

Since $900 \, \text{nm}$ falls in the infrared region, the transition must belong to the Paschen series.

For $n_1 = 3$ (Paschen series) and $n_2 = \infty$:

$$\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right),$$

$$\frac{1}{\lambda} = R \left( \frac{1}{9} - 0 \right),$$

$$\lambda = \frac{1}{R \cdot \frac{1}{9}} = \frac{9}{R}.$$λ

Substitute $R = 1.097 \times 10^7 \, \text{m}^{-1}$

$$\lambda = \frac{9}{1.097 \times 10^7} = 8.2 \times 10^{-7} \, \text{m} = 820 \, \text{nm}.$$

The transition $n_2 = \infty \to n_1 = 3$ in the Paschen series matches the infrared wavelength range close to $900 \, \text{nm}$.

Correct Answer: (d) Paschen: $\infty \to 3$

In the Bohr model of the hydrogen atom, the radius of the $n^{th}$nth orbit is given by:

$$r_n = n^2 a_0$$​

where $a_0$​ is the Bohr radius (a constant).
$$\boxed{r_n \propto n^2}$$​

From Bohr's model, energy of hydrogen-like species is:

\[
E_n = -13.6 \cdot \frac{Z^2}{n^2} \, \text{eV}
\]

**(A)**
H atom ground state: \( Z = 1, n = 1 \Rightarrow E = -13.6 \, \text{eV} \)
He\(^+\) first excited state: \( Z = 2, n = 2 \Rightarrow E = -13.6 \cdot \frac{4}{4} = -13.6 \, \text{eV} \) ✅

**(B)**
H ground: \( -13.6 \, \text{eV} \)
Li\(^{++}\) second excited state: \( Z = 3, n = 3 \Rightarrow E = -13.6 \cdot \frac{9}{9} = -13.6 \, \text{eV} \) ✅

**(C)**
H ground: \( -13.6 \, \text{eV} \)
He\(^+\) ground: \( Z = 2, n = 1 \Rightarrow E = -13.6 \cdot 4 = -54.4 \, \text{eV} \) ❌

**(D)**
He\(^+\) first excited: \( Z = 2, n = 2 \Rightarrow E = -13.6 \cdot \frac{4}{4} = -13.6 \, \text{eV} \)
Li\(^{++}\) ground: \( Z = 3, n = 1 \Rightarrow E = -13.6 \cdot 9 = -122.4 \, \text{eV} \) ❌

Correct statements: **(A), (B)**

For hydrogen atom, the Balmer series corresponds to transitions ending at $n_2 =2$
Thus, the general transition is $n_i\to 2, where n_i =3,4,5...$
The first Balmer line $\left(H_{\alpha }\right)$ : $n_i =3\to 2$
The second Balmer line ($H_{\beta }$) : $n_i =4 \to 2$
Hence, the transition giving the second Balmer line is $n_i =4 \to n_f =2$

for Balmer series limit, $\frac{1}{{\lambda }_B}=R\left[\frac{1}{{\left(2\right)}^2}-\frac{1}{{\infty }^2}\right]=\frac{R}{4}$

for Lyman series limit, $\frac{1}{{\lambda }_l}=k\left[\frac{1}{{\left(1\right)}^2}-\frac{1}{{\infty }^2}\right]=R$

$\left(\frac{{\lambda }_B}{{\lambda }_l}=4\right)$

for second line of Balmer series, ${\mathrm{n}}_{\mathrm{f}} = 2, {\mathrm{n}}_{\mathrm{i}} = 4$

In the alpha particle scattering experiment, the impact parameter is the perpendicular distance between the path of an alpha particle and the center of the nucleus. The scattering angle is the angle at which the alpha particle is deflected. The correct statements are as follows:

(A) For a scattering angle \(\theta \approx 0\), the impact parameter is large because the alpha particle passes far from the nucleus and experiences a small deflection.

(B) For a scattering angle \(\theta \approx \pi\), the impact parameter is small because the alpha particle passes very close to the nucleus and experiences a large deflection.

(C) The number of alpha particles undergoing head-on collision is small because head-on collisions are rare events.

(D) The experiment provides an estimate of the upper limit to the size of the target nucleus because the closest approach of the alpha particles gives information about the size of the nucleus.

For hydrogen atom, the Balmer series corresponds to transitions ending at $n_2 =2$
Thus, the general transition is $n_i\to 2, where n_i =3,4,5...$
The first Balmer line $\left(H_{\alpha }\right)$ : $n_i =3\to 2$
The second Balmer line ($H_{\beta }$) : $n_i =4 \to 2$
Hence, the transition giving the second Balmer line is $n_i =4 \to n_f =2$