Question

An isotope of radium (²²⁴₈₈Ra) undergoes alpha decay. Use the information below to determine the deBroglie wavelength of the emitted alpha particle.

Atomic mass ²²⁴₈₈Ra: 224.020187 u

m_alpha: 4.0026 u (6.646 x 10^-27 kg)

Atomic number	Element	Mass number	Atomic mass (u)
90	Thorium (Th)	228	228.028716
		226	225.905364
88	Radium (Ra)	220	220.043941
86	Radon (Rn)	222	222.017571
		220	220.011369

Answer 1

The alpha decay of ²²⁴₈₈Ra can be represented as,

²²⁴₈₈Ra →     ²²⁰₈₆Rn + ⁴₂He

The Q value of the reaction is

Q = {mass of ²²⁴₈₈Ra – (mass of ²²⁰₈₆Rn + mass of ⁴₂He)} x 931.5 MeV

Q = {224.020187 – (220.011369 + 4.0026)} x 931.5 = 5.792 MeV

Kinetic energy of alpha particle emitted, K_α = mass of daughter nucleus x Q/ (mass of alpha particle+ mass of daughter nucleus)

K_α = 220 x Q/224 = 5.794 x 220/224 = 5.69 MeV = 5.69 x 1.6 x 10^-13 J = 9.104 x 10^-13 J

De-Broglie wavelength of alpha particle, λ = h/(2m K_α)^1/2

λ = 6.625 x 10^-34/( 2 x 6.646 x 10^-27 x 9.104 x 10^-13)^1/2 = 6.022 x 10^-13 m

De-Broglie wavelength of alpha particle, λ = 6.022 x 10^-13 m