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The radionuclide .^(238)U decays by emit...

The radionuclide `.^(238)U` decays by emitting an alpha particle. `.^(238)Uto.^(234)Th+.^(4)He` The atomic masses of the three isotopes are `.^(238)" "U 238.05079 am u`
`.^(234) U" "234.040363 am u`
`.^(4) He " " 4.00260 am u` What is the maximum kinetic energy of the emitted alpha particle. Express your answer in Joule. (`1am u = 1.67 xx 10^(-27)kg`)

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We are given the following atomic masses: ""_(93)Pu^(238) = 238.04954 u ""_(92)U^(234) = 234.04096 u ""_(2)He^(4) = 4.00260 u Calculate the kinetic energy associated with the alpha particle emitted during the conversion of ""_(94)Pu^(238) into ""_(92)U^(234)

._(92)^(238)U has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes:

._(92)^(238)U has 92 protons and 238 nucleons. It decays by emitting an alpha particle and becomes:

Find the kinetic energy of the alpha - particle emitted in the decay ^238 Pu rarr ^234 U + alpha . The atomic masses needed are as following: ^238 Pu 238.04955 u ^234 U 234.04095 u ^4 He 4.002603 u . Neglect any recoil of the residual nucleus.

Find the kinetic energy of the alpha - particle emitted in the decay ^238 Pu rarr ^234 U + alpha . The atomic masses needed are as following: ^238 Pu 238.04955 u ^234 U 234.04095 u ^4 He 4.002603 u . Neglect any recoil of the residual nucleus.

Find the kinetic energy of the alpha - particle emitted in the decay ^238 Pu rarr ^234 U + alpha . The atomic masses needed are as following: ^238 Pu 238.04955 u ^234 U 234.04095 u ^4 He 4.002603 u . Neglect any recoil of the residual nucleus.

which a U^(238) nucleus original at rest , decay by emitting an alpha particle having a speed u , the recoil speed of the residual nucleus is