Atoms Previous Year Questions — JEE Main Physics
22 past-year Atoms questions from JEE Main (Physics) — page 3.
Q53 (2024)
Hydrogen atom is bombarded with electrons accelerated through a potential difference of $V$, which causes excitation of hydrogen atoms. If the experiment is being performed at $T = 0\text{ K}$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{\alpha}{10}\text{ V}$, where $\alpha =$ ________.
### Related Formula
The energy of an electron in the $n$-th state of a hydrogen atom is:
$$E_n = -\frac{13.6}{n^2}\text{ eV}$$
### Core Logic
At $T = 0\text{ K}$, all hydrogen atoms are in their ground state ($n = 1$).
To observe any lines in the Balmer series emission spectrum:
* The emission lines must result from a transition terminating in the state $n = 2$.
* This requires the hydrogen electron to first be excited to at least the state $n = 3$.
Thus, the bombarding electrons must have enough kinetic energy to excite the ground-state electron ($n = 1$) to the $n = 3$ state.
### Step 1: Calculate the Required Energy and Potential
The required energy difference is:
$$\Delta E = E_3 - E_1 = -\frac{13.6}{3^2} - \left(-\frac{13.6}{1^2}\right)\text{ eV}$$
$$\Delta E = 13.6 \left(1 - \frac{1}{9}\right)\text{ eV} = 13.6 \times \frac{8}{9}\text{ eV} \approx 12.09\text{ eV}$$
Since the excitation energy is provided by electrons accelerated through potential $V$, the minimum potential difference required is:
$$V \approx 12.09\text{ V} \approx 12.1\text{ V}$$
### Step 2: Solve for Alpha
Comparing to the given form $\frac{\alpha}{10}\text{ V}$:
$$12.1 = \frac{\alpha}{10} \implies \alpha = 121$$
### Pattern Recognition
Balmer series emission requires excitation of the electron to at least $n=3$. Ground state $n=1$ to $n=3$ excitation energy is $12.09\text{ eV} \approx 12.1\text{ eV}$, matching $\alpha = 121$ when multiplied by $10$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Physics: Atoms
Q41 (2024)
The radius of the third stationary orbit of an electron for a Bohr's atom is $R$. The radius of the fourth stationary orbit will be:
- $\frac{4}{3}R$
- $\frac{16}{9}R$
- $\frac{3}{4}R$
- $\frac{9}{16}R$
### Related Formula
$$r_n \propto \frac{n^2}{Z}$$
Where $n$ is the principal quantum number and $Z$ is the atomic number.
### Core Logic
For a given atom, $Z$ is constant, hence:
$$\frac{r_4}{r_3} = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$$
Given $r_3 = R$:
### Step 1: Calculate the final radius
$$r_4 = \frac{16}{9}R$$
### Pattern Recognition
Bohr orbital dimensions scale quadratically with quantum index ($n^2$), rendering quick ratios straightforward via simple squaring operations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q1641 (2024)
When a hydrogen atom going from $n = 2$ to $n = 1$ emits a photon, its recoil speed is $\frac{x}{5} \mathrm{~m / s}$. Where $x =$ _______. (Use: mass of hydrogen atom $= 1.6 imes 10^{-27} \mathrm{~kg}$)
### Related Formula
By conservation of linear momentum, the momentum of the recoiling hydrogen atom matches the momentum of the emitted photon:
$$p_{\text{atom}} = p_{\text{photon}} \implies m v = \frac{\Delta E}{c}$$
Hence, the recoil speed $v$ is:
$$v = \frac{\Delta E}{m c}$$
where $\Delta E$ is the energy difference between the electronic energy levels.
### Core Logic
For a hydrogen transition from $n = 2$ to $n = 1$:
$$\Delta E = E_2 - E_1 = -3.4 \mathrm{~eV} - (-13.6 \mathrm{~eV}) = 10.2 \mathrm{~eV}$$
Converting energy to Joules:
$$\Delta E = 10.2 \times 1.6 imes 10^{-19} \mathrm{~J}$$
{{SOL_IMG1}}
### Step 1: Compute Recoil Speed
Substituting given constant values:
$$v = \frac{10.2 \times 1.6 imes 10^{-19} \mathrm{~J}}{(1.6 imes 10^{-27} \mathrm{~kg}) \times (3 imes 10^{8} \mathrm{~m/s})}$$
### Step 2: Simplify Arithmetic Fractions
Cancelling out the factor $1.6$ from numerator and denominator:
$$v = \frac{10.2 \times 10^{-19}}{3 imes 10^{-19}} = \frac{10.2}{3} = 3.4 \mathrm{~m/s}$$
### Step 3: Solve for Variable x
Equating the computed speed to the given parameter form:
$$3.4 = \frac{x}{5} \implies x = 3.4 \times 5 = 17$$
Therefore, the value of $x$ is $17$.
### Pattern Recognition
Always look for clean component cancellation matches prior to raw exponential expansion. In this problem, the value $1.6$ in the electron charge value cancels with the structural atomic mass constant value perfectly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q37 (2024)
An electron revolving in $n^{\text{th}}$ Bohr orbit has magnetic moment $\mu_{\mathrm{n}}$. If $\mu_{\mathrm{n}} \propto n^{\mathrm{x}}$, the value of $\mathbf{x}$ is:
- $2$
- $1$
- $3$
- $0$
### Related Formula
$$\mu_n = i A = \frac{e}{T} \pi r^2$$
Alternative relation using angular momentum:
$$\mu_n = \frac{e}{2m} L$$
### Core Logic
From Bohr's quantization condition, angular momentum $L$ is given by:
$$L = \frac{nh}{2\pi}$$
Therefore, the magnetic moment can be written directly as:
$$\mu_n = \frac{e}{2m} \left(\frac{nh}{2\pi}\right)$$
$$\implies \mu_n \propto n$$
### Step 1: Direct Proportionality Analysis
We can also derive this via velocity and radius:
$v \propto \frac{1}{n}$ and $r \propto n^2$.
$$\mu_n = \frac{e v r}{2} \propto \left(\frac{1}{n}\right) (n^2) = n$$
Comparing with $\mu_n \propto n^x$, we find $x = 1$.
### Pattern Recognition
Magnetic moment of an electron in a Bohr orbit is always directly proportional to the principal quantum number $n$. It is an integer multiple of the Bohr magneton ($\mu_B = \frac{eh}{4\pi m}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
Q36 (2024)
The ratio of the magnitude of the kinetic energy to the potential energy of an electron in the $5^{\text{th}}$ excited state of a hydrogen atom is :
- $4$
- $\frac{1}{4}$
- $\frac{1}{2}$
- $1$
### Related Formula
$\text{KE} = -E$
$\text{PE} = 2E$
$$\text{KE} = \frac{1}{2} |\text{PE}|$$
### Core Logic
In any allowed Bohr orbit (for any value of principal quantum number $n$), the relationship between kinetic energy (KE), potential energy (PE), and total energy (E) strictly obeys the virial theorem for a Coulombic force field.
$$|\text{PE}| = 2 \times \text{KE}$$
### Step 1: Formulate the Ratio
The question asks for the ratio of the magnitude of KE to the magnitude of PE.
$$\frac{\text{KE}}{|\text{PE}|} = \frac{1}{2}$$
This ratio is independent of the orbit state. Even though it is the $5^{\text{th}}$ excited state, the ratio remains $\frac{1}{2}$.
### Pattern Recognition
Energy relationships in Bohr orbits (and planetary motion): $K = -E = -U/2$. The state number ($n$) is given solely as a distractor.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Atoms
