When U238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

Question

When U238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is (A) (4u/238) (B) -(4u/234) (C) (4u/234) (D) -(4u/238)

Tardigrade Admin · Accepted Answer

According to principle of conservation of linear momentum the momentum of the system remains the same before and after the decay.
Atomic mass of uranium = 238 and after emitting an alpha particle

= 238 - 4 = 234

∴ 238 \times 0 = 4 u + 234 v

∴ v = - \frac{4 u}{234}

Q. When $U^{238}$ nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

$\frac{4 u}{238}$

$- \frac{4 u}{234}$

$\frac{4 u}{234}$

$- \frac{4 u}{238}$

According to principle of conservation of linear momentum the momentum of the system remains the same before and after the decay.
Atomic mass of uranium = 238 and after emitting an alpha particle
$= 238 - 4 = 234$
$∴ 238 \times 0 = 4 u + 234 v$
$∴ v = - \frac{4 u}{234}$

Q. When U238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

2384u​

−2344u​

2344u​

−2384u​