Board-style MCQs and past-paper numericals in the BIEK / Sindh Board pattern. Tap an option to check yourself instantly. Solved questions are at the bottom.
Multiple-choice questions
Solved numericals (past papers)
Energy of a photon (E = hf)
Find the energy of a photon of light of frequency 5.0 × 10¹⁴ Hz (h = 6.63 × 10⁻³⁴ J·s).
E = hf = (6.63 × 10⁻³⁴)(5.0 × 10¹⁴)
= 3.32 × 10⁻¹⁹ J ≈ 2.07 eV
Photon energy from wavelength
Find the energy of a photon of wavelength 500 nm (h = 6.63 × 10⁻³⁴ J·s, c = 3.0 × 10⁸ m/s).
E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / (500 × 10⁻⁹)
= (1.989 × 10⁻²⁵) / (5.0 × 10⁻⁷) = 3.98 × 10⁻¹⁹ J ≈ 2.49 eV
Maximum kinetic energy of a photoelectron
Light of energy 3.5 eV strikes a metal of work function 2.0 eV. Find the maximum K.E. of the emitted electrons.
K.E.(max) = hf − φ = 3.5 − 2.0
= 1.5 eV (= 2.4 × 10⁻¹⁹ J)
Threshold frequency from work function
A metal has a work function of 3.31 × 10⁻¹⁹ J. Find its threshold frequency (h = 6.63 × 10⁻³⁴ J·s).
f₀ = φ/h = (3.31 × 10⁻¹⁹) / (6.63 × 10⁻³⁴)
= 5.0 × 10¹⁴ Hz
Stopping potential
The maximum K.E. of a photoelectron is 2.0 eV. Find the stopping potential V₀.
eV₀ = K.E.(max) ⟹ V₀ = K.E.(max)/e = 2.0 eV / e
= 2.0 V
de Broglie wavelength of an electron
An electron (m = 9.1 × 10⁻³¹ kg) moves at 7.3 × 10⁶ m/s. Find its de Broglie wavelength (h = 6.63 × 10⁻³⁴ J·s).
λ = h/(mv) = (6.63 × 10⁻³⁴) / (9.1 × 10⁻³¹ × 7.3 × 10⁶)
= (6.63 × 10⁻³⁴) / (6.64 × 10⁻²⁴) = 1.0 × 10⁻¹⁰ m (0.1 nm)