Two positively charged particles m1 and m2 have been accelerated across the same potential difference of 200 keV as…

Year	2026	2025	2024	Total
Questions	14	16	9	39

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

λ = h\sqrt(2m(K.E.))

where K.E. is the kinetic energy acquired, which equals $qV$ $qV$.

Core Logic

Since both particles have the same charge and are accelerated through the same potential difference (200 keV), they will acquire the exact same kinetic energy. So, $λ \propto 1\sqrt(m)$ $\lambda \propto \frac{1}{\sqrt{m}}$

Ratio of their de Broglie wavelengths:

(λd)m₁(λd)m₂ = \sqrt((m₂)/(m₁)) = \sqrt((4)/(1)) = 2

Therefore, $(λd)m₁ = 2 (λd)m₂$

Step 1: Final Conclusion

The value of $x$ $x$ is 2.

Pattern Recognition

For same K.E., $λ$ $\lambda$ depends purely on the inverse square root of mass.

Chapter Mix

Class 11 Chemistry: Structure of Atom Class 12 Physics: Dual Nature of Radiation and Matter

More Structure of Atom Previous-Year Questions — Page 8

Q90 jee_main_2024_30_january_evening Atomic Spectra

Number of spectral lines obtained in

He^+

spectra, when an electron makes transition from fifth excited state to first excited state will be

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

Number of spectral lines = (Δ n (Δ n + 1))/(2)

where $Δ n = n₂ - n₁$ $\Delta n = n_2 - n_1$

Core Logic

Identify the principal quantum numbers ( $n$ $n$) for the given states:

Fifth excited state means $n₂ = 5 + 1 = 6$ $n_2 = 5 + 1 = 6$.
First excited state means $n₁ = 1 + 1 = 2$ $n_1 = 1 + 1 = 2$.

Calculate the difference:

Δ n = n₂ - n₁ = 6 - 2 = 4

Step 1: Calculate Total Spectral Lines

Substitute $Δ n = 4$ $\Delta n = 4$ into the formula:

Maximum number of spectral lines = (4(4 + 1))/(2)

= (4 \times 5)/(2)

$= 10$ $= 10$

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q68 jee_main_2024_30_jan_morning Quantum Mechanical Model

Given below are two statements: Statement-I: The orbitals having same energy are called as degenerate orbitals. Statement-II: In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals. In the light of the above statements, choose the most appropriate answer from the options given

A. $Statement-I is true but Statement-II is false$
B. $Both Statement-I and Statement-II are true.$
C. $Both Statement-I and Statement-II are false$
D. $Statement-I is false but Statement-II is true$

Solution

Core Logic

Statement-I defines degenerate orbitals correctly: Orbitals that share the exact same energy level are called degenerate orbitals. Statement-II claims 3p and 3d are not degenerate in a hydrogen atom. For a single-electron system like Hydrogen ( $1s¹$ $1s^1$), the energy of an orbital depends only on the principal quantum number ' $n$ $n$'.

Step 1: Final conclusion

In a hydrogen atom, energy of $3s = 3p = 3d$ $3s = 3p = 3d$. Therefore, 3p and 3d are degenerate. Thus, Statement-II is false.

Pattern Recognition

Hydrogen atom (single electron species) = Energy dictated strictly by $n$ $n$. Multi-electron atoms = Energy dictated by $(n + l)$ $(n + l)$ rule.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q jee_main_2024_31_jan_evening Quantum Numbers

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

A. $n = 4, l = 2, m = -1, s = +(1)/(2)$
B. $n = 4, l = 0, m = 0, s = +(1)/(2)$
C. $n = 3, l = 0, m = 1, s = +(1)/(2)$
D. $n = 2, l = 0, m = 0, s = +(1)/(2)$

Solution

Core Logic

The atomic number of potassium (K) is $19$ $19$. Its electronic configuration is: $1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹$ $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost orbital of potassium is the $4s$ $4s$ orbital.

For the $4s$ $4s$ orbital: Principal quantum number, $n = 4$ $n = 4$ Azimuthal quantum number, $l = 0$ $l = 0$ (for s-subshell) Magnetic quantum number, $m = 0$ $m = 0$ (since $l = 0$ $l = 0$) Spin quantum number, $s = \pm (1)/(2)$ $s = \pm \frac{1}{2}$

Step 1: Final Matching

Comparing these values with the options, option (2) provides $n = 4, l = 0, m = 0, s = +(1)/(2)$ $n = 4, l = 0, m = 0, s = +\frac{1}{2}$, which perfectly describes the outermost electron.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q89 jee_main_2024_31_jan_morning Photoelectric Effect and Ionization Energy

The ionization energy of sodium in

kJ mol⁻¹

. If electromagnetic radiation of wavelength 242 nm is just sufficient to ionize sodium atom is ________

Numerical Answer. Answer: 493 to 495

Solution

Related Formula

E = 1240λ (nm) eV

Etotal = E \times NA

Step 1: Energy per atom

E = (1240)/(242) eV = 5.12 eV

Converting eV to Joules:

E = 5.12 \times 1.6 \times 10⁻¹⁹ J/atom

E = 8.192 \times 10⁻¹⁹ J/atom

Step 2: Energy per mole

To find the ionization energy per mole, multiply by Avogadro's number ( $NA \approx 6.022 \times 10²³$ $N_A \approx 6.022 \times 10^{23}$):

IE = 8.192 \times 10⁻¹⁹ \times 6.022 \times 10²³ J/mol

IE \approx 493.3 \times 10³ J/mol \approx 494 kJ/mol

Chapter Mix

Class 11 Chemistry: Structure of Atom

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Total Spectral Lines

Chapter Mix

Solution

Core Logic

Step 1: Final conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Matching

Chapter Mix

Solution

Related Formula

Step 1: Energy per atom

Step 2: Energy per mole

Chapter Mix

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More Structure of Atom Previous-Year Questions — Page 8

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Core Logic

Step 1: Final conclusion

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Step 1: Final Matching

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Solution

Related Formula

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