A nucleus ⁹⁴₃₈Sr undergoes beta-minus decay followed by another beta-minus decay. The final nucleus is:
Answer: A
First β⁻ decay: ⁹⁴₃₈Sr → ⁹⁴₃₉Y + e⁻ + ν̄. Second β⁻ decay: ⁹⁴₃₉Y → ⁹⁴₄₀Zr + e⁻ + ν̄. Final nucleus is ⁹⁴₄₀Zr.
Q.33Medium
A photon of frequency 6 × 10¹⁵ Hz is incident on a metal surface with work function 2 eV. What is the maximum kinetic energy of the ejected photoelectron? (h = 4.14 × 10⁻¹⁵ eV·s)
Answer: B
Energy of photon E = hf = 4.14 × 10⁻¹⁵ × 6 × 10¹⁵ = 24.84 eV. Maximum KE = E - W = 24.84 - 2 = 22.84 eV (approximately 21.84 eV with standard constants).
Q.34Medium
An electron in the first excited state of hydrogen atom (n=2) transitions to ground state (n=1). The wavelength of emitted photon is approximately:
Answer: A
Using Rydberg formula: 1/λ = R(11² - 21²) = R(43). With R = 1.097 × 10⁷ m⁻¹, λ ≈ 121.6 nm (Lyman alpha line).
Q.35Medium
The de Broglie wavelength of an electron with kinetic energy 50 eV is:
Answer: A
λ = h/p = h/√(2mE). For 50 eV electron: λ = 6.63×10⁻³⁴/√(2×9.1×10⁻³¹×50×1.6×10⁻¹⁹) ≈ 0.173 nm.
Q.36Medium
In Compton scattering, a photon collides with a free electron at rest. Which quantity always increases?
Answer: B
In Compton effect, photon transfers energy to electron, losing energy and increasing wavelength. Δλ = (h/m_e c)(1 - cosθ).
Q.37Medium
The activity of a radioactive sample decreases from 8000 Bq to 1000 Bq in 20 hours. The half-life of the sample is:
Answer: A
A = A₀(21)^(t/T₁/₂). 1000 = 8000(21)^(20/T₁/₂). (81) = (21)³, so 20/T₁/₂ = 3, T₁/₂ = 6.67 hours.
Q.38Medium
In a nuclear reactor, control rods are used to absorb neutrons. Which isotope is commonly used in control rods?
Answer: A
Boron-10 and Cadmium have high neutron absorption cross-sections and are used in control rods to regulate chain reactions in nuclear reactors.
Q.39Medium
A radioactive nucleus ₆₀₂₇Co undergoes beta-plus decay. The daughter nucleus is:
Answer: A
In beta-plus decay, Z decreases by 1, A remains constant. ⁶⁰₂₇Co → ⁶⁰₂₆Ni + e⁺ + νₑ.
Q.40Medium
The threshold frequency for photoelectric effect in a metal is 6 × 10¹⁴ Hz. The work function of the metal is:
Answer: A
Work function W = hf₀ = 6.63 × 10⁻³⁴ × 6 × 10¹⁴ = 3.98 × 10⁻¹⁹ J ≈ 2.48 eV.