Practice Pakistan questions with answers and explanations.
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Photon energy is E=hf.
E=6.63×10^-34×5×10^14 = 3.315×10^-19 J.
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Boyle’s law gives P₁V₁ = P₂V₂.
P₂ = 120×3÷6 = 60 kPa.
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Frequency is f=E/h.
F=4.0×10^-19 / 6.63×10^-34 ≈ 6.03×10^14 Hz.
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Boyle’s law gives P₁V₁ = P₂V₂.
P₂ = 80×5÷4 = 100 kPa.
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Einstein's photoelectric equation gives Kmax=Ephoton-φ.
Kmax=5-2 = 3 eV.
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Boyle’s law gives P₁V₁ = P₂V₂.
P₂ = 150×2÷5 = 60 kPa.
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Use λ=h/p.
λ=6.63×10^-34 / 6.63×10^-24 = 1×10^-10 m.
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Boyle’s law gives P₁V₁ = P₂V₂.
P₂ = 90×6÷3 = 180 kPa.
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The photon energy equals the difference between levels.
E=5-2 = 3 eV.
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Voltage ratio equals turns ratio.
Vs=20×500/100 = 100 V.
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One half-life reduces the undecayed population by half.
800/2 = 400 nuclei.
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A photon carries energy proportional to frequency.
The photon model explains effects such as photoelectric emission.
