Correct answer option is (D)

To determine how many spectral lines will be emitted due to transitions of electrons in a hydrogen atom when it is given an energy of $10.2\text{eV}$, we first need to ascertain which energy level the electron will reach with this energy and then count the possible transitions (spectral lines) as it returns to the ground state.

Energy Levels of Hydrogen Atom :

The energy levels $E_n$ of a hydrogen atom can be calculated using the formula:

$E_n=-\frac{13.6\text{eV}}{n^2}$

where $n$ is the principal quantum number.

Ground State Energy :

The ground state (n=1) energy is $E_1=-13.6\text{eV}$.

Determine the Excited State :

If the ground state electron is given $10.2\text{eV}$, its total energy becomes:

$E_{\text{total}}=E_1+10.2\text{eV}=-13.6\text{eV}+10.2\text{eV}=-3.4\text{eV}$

Now, we find the principal quantum number $n$ for which the energy is closest to $-3.4\text{eV}$:

• For $n=2$: $E_2=-\frac{13.6}{2^2}=-3.4\text{eV}$


• For $n=3$: $E_3=-\frac{13.6}{3^2}=-1.51\text{eV}$

Since $-3.4\text{eV}$ matches exactly with $E_2$, the electron reaches the second energy level ($n=2$).

Counting Spectral Lines :

When the electron falls back to the ground state from $n=2$, it can do so in a single transition:

• $n=2$ to $n=1$

Thus, only one spectral line will be emitted during this transition.

Conclusion :

The number of spectral lines emitted when a hydrogen atom in the ground state is given $10.2\text{eV}$ and the electron transitions back to the ground state from $n=2$ is just one.

Therefore, the correct answer is:

Option D: 1.

Correct answer option is (B)

To determine the longest wavelength associated with the Paschen series, we need to understand what the Paschen series is and how its wavelength can be calculated. The Paschen series pertains to the spectral line emissions of the hydrogen atom as an electron transitions from higher energy levels (n > 3) down to n = 3. The formula used to calculate the wavelength ($\lambda$) of the emitted photon during such a transition in the hydrogen atom is given by the Rydberg formula:

$\frac{1}{\lambda }=R_H(\frac{1}{3^2}-\frac{1}{n^2})$

where,

• $R_H$ is the Rydberg constant for hydrogen, $R_H=1.097\times {10}^7{\text{m}}^{-1}$.
• $n$ is the principal quantum number of the higher energy level (For the longest wavelength, we use the smallest possible value of $n$ that is larger than 3, which is $n=4$, since for longer wavelength transitions, the difference in energy levels should be the smallest).

To find the longest wavelength, we plug in $n=4$ into the Rydberg equation, as this corresponds to the smallest energy transition within the Paschen series ($n=4$ to $n=3$):

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{3^2}-\frac{1}{4^2})$

$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{9}-\frac{1}{16})$

$\frac{1}{\lambda }=1.097\times {10}^7\left(\frac{16-9}{144}\right)$

$\frac{1}{\lambda }=1.097\times {10}^7\times \frac{7}{144}$

$\frac{1}{\lambda }=1.097\times {10}^7\times \frac{7}{144}$

$\frac{1}{\lambda }=1.097\times {10}^7\times \frac{7}{144}$

$\frac{1}{\lambda }=1.097\times {10}^7\times \frac{7}{144}=76403.47{\text{m}}^{-1}$

Finally, we calculate $\lambda$ by taking the reciprocal of this value:

$\lambda =\frac{1}{76403.47}\approx 1.308\times {10}^{-5}\text{m}=1.308\times {10}^{-5}\times {10}^2\text{cm}=1.308\times {10}^{-3}\text{cm}$

It seems there was a mistake in my calculation. Revising the calculation properly:

$\lambda =\frac{1}{1.097\times {10}^7\times \frac{7}{144}}$

Correctly evaluating this, we should directly compute:

$\lambda \approx \frac{144}{7\times 1.097\times {10}^7}=\frac{144}{7.679\times {10}^7}\approx 1.876\times {10}^{-6}\text{m}$

Therefore, the correct option that matches our calculation for the longest wavelength in the Paschen series is:

Option B: $1.876\times {10}^{-6}\text{m}$.

Correct answer option is (D)

The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:

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

where:

• $\lambda$ is the wavelength of the emitted light,
• $R$ is the Rydberg constant,
• $n_1$ is the lower energy level, and
• $n_2$ is the higher energy level.

The Balmer series corresponds to electron transitions where the lower energy level, $n_1$, is 2, and the Lyman series corresponds to transitions where $n_1$ is 1. The shortest wavelength in each series occurs for transitions from the highest possible energy level ($n_2=\mathrm{\infty }$) to the specified lower level ($n_1$).

For the Balmer series (shortest wavelength):

$n_1=2,n_2=\mathrm{\infty }$,

Putting these values into the Rydberg formula gives:

$\frac{1}{{\lambda }_B}=R(\frac{1}{2^2}-\frac{1}{{\mathrm{\infty }}^2})$

$\frac{1}{{\lambda }_B}=R\left(\frac{1}{4}\right)$

${\lambda }_B=\frac{1}{R\cdot \frac{1}{4}}=\frac{4}{R}$

For the Lyman series (shortest wavelength):

$n_1=1,n_2=\mathrm{\infty }$,

Putting these values into the Rydberg formula gives:

$\frac{1}{{\lambda }_L}=R(\frac{1}{1^2}-\frac{1}{{\mathrm{\infty }}^2})$

$\frac{1}{{\lambda }_L}=R\cdot 1$

${\lambda }_L=\frac{1}{R}$

The ratio of the shortest wavelength of Balmer series (${\lambda }_B$) to the shortest wavelength of Lyman series (${\lambda }_L$) is:

$\frac{{\lambda }_B}{{\lambda }_L}=\frac{\frac{4}{R}}{\frac{1}{R}}=\frac{4}{1}=4:1$

Therefore, the correct answer is Option D: $4:1$.

Correct answer option is (C)

The energy $E_n$ of an electron in the $n$th energy level of a hydrogen atom is given by the formula:

$E_n=-\frac{13.6\text{eV}}{n^2}$

where:

• $E_n$ is the energy in electron volts (eV),


• $13.6\text{eV}$ is the ionization energy of hydrogen (the energy required to remove the electron from the ground state, $n=1$),


• $n$ is the principal quantum number (an integer starting from 1).

The Balmer Series

The Balmer series is a specific set of spectral lines of hydrogen that are visible to the naked eye. It results from the electron making transitions between energy levels, specifically from levels with $n\ge 3$ (i.e., from higher energy states) down to $n=2$. These transitions release energy in the form of electromagnetic radiation, which we observe as the Balmer series.

1. Ground State Energy ($E_1$): The energy of the electron in the ground state ($n=1$) of hydrogen is:

$E_1=-\frac{13.6\text{eV}}{1^2}=-13.6\text{eV}$

This represents the electron's energy level when it's closest to the nucleus.

1. Energy for the $n=3$ Level ($E_3$): For an electron in the $n=3$ level, its energy is:

$E_3=-\frac{13.6\text{eV}}{3^2}=-1.51\text{eV}$

1. Minimum Energy for Balmer Series Transition: The minimum energy transition within the Balmer series is from $n=3$ to $n=2$, but to understand the total energy required for an electron to emit radiation in the Balmer series, starting from the ground state, we consider the energy needed to first excite the electron from $n=1$ to $n=3$.


2. Energy Difference ($\mathrm{\Delta }E$): The energy emitted as radiation for the electron to transition from the ground state to a state where it can participate in the Balmer series is the difference between the ground state energy and the energy of the $n=3$ state:

$\mathrm{\Delta }E=E_3-E_1=-1.51\text{eV}-(-13.6\text{eV})=12.09\text{eV}$

Note

This calculation shows that the minimum energy required by a hydrogen atom in the ground state to emit radiation in the Balmer series is approximately 12.1 eV. The energy difference essentially represents the energy that must be supplied to an electron to excite it from the ground state ($n=1$) to an excited state ($n=3$) from which it can then transition to $n=2$, emitting radiation observable as part of the Balmer series.

Correct answer option is (C)

$\begin{array}{rl}& \frac{1}{\lambda }=\frac{13.6{\mathrm{z}}^2}{\mathrm{hc}}[\frac{1}{1^2}-\frac{1}{2^2}]......\text{(i)}\\ & \frac{1}{{\lambda }^{\prime }}=\frac{13.6{\mathrm{z}}^2}{\mathrm{hc}}[\frac{1}{1^2}-\frac{1}{3^2}]......\text{(ii)}\end{array}$

On dividing (i) & (ii)

${\lambda }^{\prime }=\frac{27}{32}\lambda$