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Conservation of energy:

${\mathit{K}}_{\mathbf{T}}\mathit{h}\mathbf{+}\mathit{Q}\mathbf{=}{\mathit{K}}_{\mathbf{R}}\mathit{a}\mathbf{+}{\mathit{K}}_{\mathbf{伪}}$

Conservation of momentum:

$\mathit{p}{\mathbf{鈨�}}_{\mathbf{T}}\mathit{h}\mathbf{=}\mathit{p}{\mathbf{鈨�}}_{\mathbf{R}}\mathit{a}\mathbf{+}\mathit{p}{\mathbf{鈨�}}_{\mathbf{伪}}\mathbf{=}\mathbf{0}$

The decay process of $\stackrel{230}{90}\mathit{T}\mathit{h}$by $\mathit{伪}$emission is given as:

$\stackrel{\mathbf{230}}{\mathbf{90}}\mathit{T}\mathit{h}\mathbf{鈫�}\stackrel{\mathbf{226}}{\mathbf{88}}\mathit{R}\mathit{a}\mathbf{+}{}_{\mathbf{2}}^{\mathbf{4}}\mathit{H}\mathit{e}$

Since this decay occurs spontaneously, it releases an energy equals the difference between the mass of the nucleus $\stackrel{\mathbf{230}}{\mathbf{90}}\mathit{T}\mathit{h}$and the sum of the masses of $\stackrel{\mathbf{226}}{\mathbf{88}}\mathit{R}\mathit{a}$and ${}_{\mathbf{2}}^{\mathbf{4}}\mathit{H}\mathit{e}$multiplied by ${\mathit{c}}^{\mathbf{2}}$.

$\mathit{Q}\mathbf{=}\mathbf{\left(}\mathit{M}\mathbf{\right(}\stackrel{\mathbf{230}}{\mathbf{90}}\mathbf{\left)}\mathit{T}\mathit{h}\mathbf{\right)}\mathbf{-}\mathit{M}\mathbf{\left(}\stackrel{\mathbf{226}}{\mathbf{88}}\mathbf{\right)}\mathit{R}\mathit{a}\mathbf{)}\mathbf{-}\mathit{M}{\mathbf{(}}_{\mathbf{2}}^{\mathbf{4}}\mathit{H}\mathit{e}\mathbf{)}\mathbf{)}\mathbf{*}\mathbf{931}\mathbf{.}\mathbf{5}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u}$

(1)

Plug the given in equation (1),

localid="1667643108801" $\mathit{Q}\mathbf{=}\mathbf{[}\mathbf{230}\mathbf{.}\mathbf{033134}\mathit{u}\mathbf{-}\mathbf{226}\mathbf{.}\mathbf{025410}\mathit{u}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{002603}\mathit{u}\mathbf{]}\mathbf{*}\mathbf{931}\mathbf{.}\mathbf{5}\mathit{M}\mathit{e}\mathit{V}\mathbf{/}\mathit{u}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{770}\mathit{M}\mathit{e}\mathit{V}$

Since the parent nucleus is at rest, its kinetic energy is zero.

After the emission, the energy released is carried by the alpha particle and the daughter nucleus as kinetic energy.

From the conservation of energy,

${\mathit{K}}_{\mathbf{T}}\mathit{h}\mathbf{+}\mathit{Q}\mathbf{=}{\mathit{K}}_{\mathbf{R}}\mathit{a}\mathbf{+}{\mathit{K}}_{\mathbf{伪}}$

$\mathit{Q}\mathbf{=}{\mathit{K}}_{\mathbf{R}}\mathit{a}\mathbf{+}{\mathit{K}}_{\mathbf{伪}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{M}}_{\mathbf{R}}\mathit{a}{\mathit{V}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{m}}_{\mathbf{伪}}{\mathit{v}}_{\mathbf{伪}}^{\mathbf{2}}$ (2)

Where V is the recoil speed of $\stackrel{226}{88}$Ra.

From the conservation of momentum,

$\mathit{p}{\mathbf{鈨�}}_{\mathbf{T}}\mathit{h}\mathbf{=}\mathit{p}{\mathbf{鈨�}}_{\mathbf{R}}\mathit{a}\mathbf{+}\mathit{p}{\mathbf{鈨�}}_{\mathbf{伪}}\mathbf{=}\mathbf{0}$

$\mathbf{鈬�}\mathit{p}{\mathbf{鈨�}}_{\mathbf{R}}\mathit{a}\mathbf{=}\mathbf{-}\mathit{p}{\mathbf{鈨�}}_{\mathbf{伪}}$

${\mathit{p}}_{\mathbf{R}}\mathit{a}\mathbf{=}{\mathit{p}}_{\mathbf{伪}}\mathbf{鈬�}{\mathit{M}}_{\mathbf{R}}\mathit{a}\mathit{V}\mathbf{=}{\mathit{m}}_{\mathbf{伪}}{\mathit{v}}_{\mathbf{伪}}$

localid="1667642480461" $\mathbf{鈬�}\mathit{V}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{伪}}{\mathbf{v}}_{\mathbf{伪}}}{{\mathbf{M}}_{\mathbf{R}}\mathbf{a}}$

Substitute in equation (2),

$\mathbf{Q}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{M}}_{\mathbf{R}}\mathbf{a}{\left(\frac{m_{伪}{v}_{伪}}{M_{R}a{)}^2}\right)}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}_{\mathbf{伪}}{\mathbf{v}}_{\mathbf{伪}}^{\mathbf{2}}\mathbf{=}{\mathbf{K}}_{\mathbf{伪}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\frac{{\mathbf{m}}_{\mathbf{伪}}^{\mathbf{2}}{\mathbf{v}}_{\mathbf{伪}}^{\mathbf{2}}}{{\mathbf{M}}_{\mathbf{R}}\mathbf{a}}$

$\mathbf{鈬�}\mathit{Q}\mathbf{=}{\mathit{m}}_{\mathbf{伪}}\mathbf{+}\frac{{\mathbf{M}}_{\mathbf{R}}\mathbf{a}}{{\mathbf{M}}_{\mathbf{R}}\mathbf{a}}\mathit{K}\mathit{伪}$

$\mathbf{鈬�}{\mathit{K}}_{\mathbf{伪}}\mathbf{=}\mathit{Q}\left(\frac{M_{R}a}{M_{R}a+m_{伪}}\right)$ (3)

Plug the values,

${\mathit{K}}_{\mathbf{伪}}\mathbf{=}\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{770}\mathbf{)}\mathbf{\left(}\frac{\mathbf{226}\mathbf{.}\mathbf{025410}}{\mathbf{226}\mathbf{.}\mathbf{025410}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{002603}}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{687}\mathit{M}\mathit{e}\mathit{V}$

Thus, the kinetic energy of the emitted $\mathit{伪}$particle is $\mathbf{4}\mathbf{.}\mathbf{687}\mathit{M}\mathit{e}\mathit{V}$.