Modern Physics – JEE Main PYQs (2026) session 1

Modern Physics – JEE Main PYQs (2026) Quiz

Important Constants:
$h = 6.63 \times 10^{-34}\ \text{J·s}$
$c = 3 \times 10^8\ \text{m/s}$
$e = 1.6 \times 10^{-19}\ \text{C}$
$k = 1.38 \times 10^{-23}\ \text{J/K}$
$1\text{ eV} = 1.6 \times 10^{-19}\ \text{J}$

Score: 0

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Q1. Two electrons are moving in orbits of two hydrogen-like atoms with speeds $3 \times 10^5$ m/s and $2.5 \times 10^5$ m/s respectively. If the radii of these orbits are nearly same then the possible order of energy states are respectively:

9 and 8
8 and 10
10 and 12
6 and 5

For hydrogen-like atom:
$v \propto \frac{1}{n}$
$\frac{v_1}{v_2} = \frac{n_2}{n_1}$
$\frac{3}{2.5} = 1.2$
Closest integer ratio → 9 and 8.

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Q2. Number of photons of equal energy emitted per second by a 6 mW laser source operating at 663 nm is:

$1\times10^{16}$
$5\times10^{16}$
$2\times10^{16}$
$5\times10^{15}$

$E = \frac{hc}{\lambda}$
$E = \frac{6.63\times10^{-34}\times3\times10^8}{663\times10^{-9}}$
$E \approx 3\times10^{-19}$ J
$N = \frac{6\times10^{-3}}{3\times10^{-19}} \approx 2\times10^{16}$

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Q4. The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly:

1784 nm
1875 nm
1217 nm
1550 nm

Using Rydberg formula:
Largest Balmer: $n=3\to2$
Largest Paschen: $n=4\to3$
Numerical evaluation → ≈ 1875 nm.

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Q5. When light of given wavelength falls on metallic surface, stopping potential is 3.2 V. If wavelength is doubled, stopping potential becomes 0.7 V. The wavelength of first light is:

$3.1\times10^{-7}$ m
$2.9\times10^{-8}$ m
$2.2\times10^{-8}$ m
$2.5\times10^{-7}$ m

Photoelectric equation:
$eV = \frac{hc}{\lambda} - \phi$
Solve two equations → $\lambda_1 \approx 3.1\times10^{-7}$ m.

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Q6. A light wave described by $E = 60[\sin(3\times10^{15}t)+\sin(12\times10^{15}t)]$ falls on metal of work function 2.8 eV. Maximum kinetic energy of ejected electron is:

5.3 eV
6.0 eV
5.9 eV
5.1 eV

Higher frequency dominates:
$E = h\omega$
$E = 6.6\times10^{-34}\times12\times10^{15}$
Convert to eV → subtract 2.8 eV
$\text{KE}_{max} \approx 5.9$ eV.

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Q7. The de Broglie wavelength of an oxygen molecule at 27°C is $x\times10^{-12}$ m. Find x.

0.4
0.7
1.0
1.6

$\lambda = \frac{h}{\sqrt{3mkT}}$
Substitute values:
$\lambda \approx 0.7\times10^{-12}$ m

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