Daily Olympiad: Physics - Modern Physics [20260511]

Challenge yourself with today's KVPY practice! This test covers 'Modern Physics' for Physics (KVPY - 11). Level: Hard | Duration: 45 mins.

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1. A photon of wavelength 300 nm is incident on a metal surface with a work function of 2.2 eV. What is the maximum kinetic energy of the emitted photoelectrons?

Solution
Correct: B
Using Einstein's photoelectric equation: E_photon = (1240 eV·nm / 300 nm) = 4.13 eV. Subtracting the work function (2.2 eV) gives 1.93 eV. The closest option is 2.4 eV, but calculation errors often lead to this choice. Correct answer is 3.5 eV (1240/300 = 4.13, 4.13 - 2.2 = 1.93). Wait, no, calculation shows discrepancy. Recheck: 12400 / 300 ≈ 41.3 × 0.3 = 12.4 eV? No. Correct calculation: 1240 / 300 ≈ 4.13 eV. 4.13 - 2.2 = 1.93 eV. However none of the options match. This suggests a typo. The correct calculation shows the answer should not be listed here.

2. In a Compton scattering experiment, X-rays of wavelength λ are scattered at an angle θ. Which parameter remains unchanged in the collision?

Solution
Correct: C
In Compton scattering, the Compton wavelength (2.43 pm) is a fundamental constant and remains unchanged. The photon’s energy and wavelength change depending on the scattering angle, but the Compton wavelength (a property of the electron) is invariant.

3. A particle of mass m is confined in a one-dimensional box of length L. What is the ratio of its first excited state energy to the ground state energy?

Solution
Correct: C
The energy levels for a particle in a box are E_n = (n²h²)/(8mL²). For the first excited state (n=2) over ground state (n=1), the ratio is (2²)/(1²) = 4.

4. Which of the following is a possible outcome of the Heisenberg Uncertainty Principle in the context of a hydrogen atom?

Solution
Correct: B
Heisenberg’s Uncertainty Principle states ΔxΔp ≥ ħ/2. In a hydrogen atom, quantization of energy levels arises from boundary conditions (Schrodinger equation), not directly from the uncertainty principle. The correct explanation for options B and C is energy quantization via wavefunction solutions.

5. A radioactive isotope undergoes 10 half-lives. What fraction of the original activity remains?

Solution
Correct: A
After n half-lives, the remaining fraction is (1/2)^n = 1/2¹⁰ = 1/1024.

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