JEE Main Atom PYQ

6.(JEE Main 2024 (Online) 5th April Evening Shift)

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $915 \overset{o}{A}$ . The longest wavelength of spectral lines in the Balmer series will be _______ $\overset{o}{A}$ .

Correct answer is 6588

$\frac{1}{915} = R_{H} (\frac{1}{1^{2}} - \frac{1}{\infty^{2}})$ (For Lyman)

$\Rightarrow \frac{1}{λ} = R_{H} (\frac{1}{2^{2}} - \frac{1}{3^{2}})$ (For Balmer)

$\Rightarrow λ = 6588$ $\overset{o}{A}$

7.(JEE Main 2024 (Online) 4th April Morning Shift)

A hydrogen atom changes its state from $n = 3$ to $n = 2$ . Due to recoil, the percentage change in the wave length of emitted light is approximately $1 \times 10^{- n}$ . The value of $n$ is _______.

[Given Rhc $= 13.6 eV, hc = 1242 eV nm, h = 6.6 \times 10^{- 34} J s$ mass of the hydrogenatom $= 1.6 \times 10^{- 27} kg$ ]

Correct answer is 7

$\begin{aligned} Δ E & = 13.6 eV (\frac{1}{4} - \frac{1}{9}) \\ = \frac{68}{36} eV = 1.89 eV \end{aligned}$

Due to recoil of hydrogen atom, the energy of emitted photon will decrease by very small amount.

So for approximate calculations,

$\begin{aligned} % charge = \frac{Δ E_{atom}}{Δ E} \times 100 \\ = \frac{\frac{{(\frac{Δ E}{C})}^{2}}{2 m}}{Δ E} \times 100 \\ = \frac{Δ E}{C^{2} \times 2 m} \times 100 \\ = \frac{1.89 \times 1.6 \times 10^{- 19} \times 100}{{(3 \times 10^{8})}^{2} \times 2 \times 1.6 \times 10^{- 27}} \\ = 1.05 \times 10^{- 7} % \\ ∴ n = 7 \end{aligned}$

8.(JEE Main 2024 (Online) 1st February Evening Shift)

A particular hydrogen-like ion emits the radiation of frequency $3 \times 10^{15} Hz$ when it makes transition from $n = 2$ to $n = 1$ . The frequency of radiation emitted in transition from $n = 3$ to $n = 1$ is $\frac{x}{9} \times 10^{15} Hz$ , when $x =$ ________ .

Correct answer is 32

The emission frequency of radiation from a hydrogen-like ion during electron transitions can be understood using the formula derived from Rydberg's equation for hydrogen-like atoms, which is given as:

$E = \frac{E_{0}}{h} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$

where:

$E$ is the energy of the emitted photon,
$E_{0}$ is the Rydberg constant for hydrogen,
$h$ is Planck's constant,
$n_{1}$ and $n_{2}$ are the principal quantum numbers for the initial and final energy levels, respectively, with $n_{2} < n_{1}$ .

Given a transition from $n = 2$ to $n = 1$ emits radiation with a frequency of $3 \times 10^{15} Hz$ , we can write:

$ν_{1} = 3 \times 10^{15} Hz$

For the transition from $n = 3$ to $n = 1$ , we can use the same principle to find the frequency of the emitted radiation, $ν_{2}$ , which will depend on the difference in energy levels involved in the transition. The frequency is directly proportional to this energy difference, so we can compare the two transitions using their respective frequencies:

$\frac{ν_{2}}{ν_{1}} = \frac{(\frac{1}{n_{1 f}^{2}} - \frac{1}{n_{1 i}^{2}})}{(\frac{1}{n_{2 f}^{2}} - \frac{1}{n_{2 i}^{2}})} = \frac{(\frac{1}{1^{2}} - \frac{1}{3^{2}})}{(\frac{1}{1^{2}} - \frac{1}{2^{2}})} = \frac{(1 - \frac{1}{9})}{(1 - \frac{1}{4})}$

After simplification:

$\frac{ν_{2}}{3 \times 10^{15}} = \frac{(\frac{8}{9})}{(\frac{3}{4})} = \frac{32}{27}$

Thus, to find the frequency $u_{2}$ for the transition from $n = 3$ to $n = 1$ :

$ν_{2} = \frac{32}{9} \times 10^{15} Hz$

Hence, in the given formula $\frac{x}{9} \times 10^{15} Hz$ for the frequency, $x$ is equal to $32$ .

9.(JEE Main 2024 (Online) 30th January Morning Shift)

A electron of hydrogen atom on an excited state is having energy $E_{n} = - 0.85 eV$ . The maximum number of allowed transitions to lower energy level is _________.

Correct answer is 6

$\begin{aligned} E_{n} = - \frac{13.6}{n^{2}} = - 0.85 \\ \Rightarrow n = 4 \end{aligned}$

No of transition

$= \frac{n (n - 1)}{2} = \frac{4 (4 - 1)}{2} = 6$

10.(JEE Main 2024 (Online) 29th January Evening Shift)

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 $T = 0 K$ , the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $\frac{α}{10} V$ , where $α =$ __________.

Correct answer is 121

For minimum potential difference electron has to make transition from $n = 3$ to $n = 2$ state but first electron has to reach to $n = 3$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $n = 3$ and $n = 1$ state.

$\begin{aligned} Δ E = 13.6 [1 - \frac{1}{3^{2}}] e = eV \\ \frac{13.6 \times 8}{9} = V \\ V = 12.09 V \approx 12.1 V \end{aligned}$

So, $α = 121$