In an alpha particle scattering experiment distance of closest approach for the α particle is 4.5 × 10⁻¹⁴ m. If target nucleus has atomic number 80, then maximum velocity of α -particle is __________ × 10⁵ m/s approximately. ( (1)/(4πε₀) = 9 × 10⁹ SI unit, mass of α particle = 6.72 × 10⁻²⁷ kg) Numerical answer type. Work it out, then open the answer.

Answer: 156. 1/2mv² = (k(2e)(Ze))/(r₀) 9×10⁹ × 160 × (1.6×10⁻¹⁹)²4.5×10⁻¹⁴ = 8.19×10⁻¹³ J v² = 2 × 8.19×10⁻¹³6.72×10⁻²⁷ = 2.44×10¹⁴ v ≈ 1.56×10⁷ = 156×10⁵ m/s. Answer: 156.

In an alpha particle scattering experiment distance of closest… | JEE Main 2024 (8 Apr, Shift 1)

Q7JEE Main 2024 · 8 APR · SHIFT 1Medium

Atomic Models & Alpha-Particle Scattering

In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14}\text{ m}$ . If target nucleus has atomic number 80, then maximum velocity of $\alpha$ -particle is __________ $\times 10^{5}\text{ m/s}$ approximately.

( $\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9}\text{ SI}$ unit, mass of $\alpha$ particle $= 6.72 \times 10^{-27}\text{ kg}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 156

$\frac{1}{2}mv^2 = \dfrac{k(2e)(Ze)}{r_0}$
$\dfrac{9\times10^9 \times 160 \times (1.6\times10^{-19})^2}{4.5\times10^{-14}} = 8.19\times10^{-13}$ J
$v^2 = \dfrac{2 \times 8.19\times10^{-13}}{6.72\times10^{-27}} = 2.44\times10^{14}$
$v \approx 1.56\times10^7 = 156\times10^5$ m/s. Answer: 156.

Report a mistake

Atomic Models & Alpha-Particle Scattering question from JEE Main 2024 (8 Apr, Shift 1)

More questions on Atomic Models & Alpha-Particle Scattering