We are given the following atomic masses: 92238U=238.05079 u , 24He=4.00260 u 90234Th=234.04363 u , 11H=1.00783 u 91237Pa=237.05121 u. Take 1 u≡931.5 MeVHere the symbol Pa is for the element protactinium (Z = 91). Mark the CORRECT statement(s)

The energy released in α-decay is given by Q=(Mu−MTh−MHe)c2Substituting the atomic masses as given in the data, we findQ=(238.05079−234.04363−4.00260)u×c2=(0.00456 u)c2=(0.00456 u)c2 =(0.00456 u)(931.5 MeV/u)=4.25 MeV .If 92238U spontaneously emits a proton, the decay process would be 92238U→ 91237Pa+ 11HThe Q for this process to happen is =(MU−MPa−MH)c2=(238.05079−237.05121−1.00783)u×c2=(−0.00825 u)c2 =−(0.00825 u)(931.5 MeV/u) =−7.68 MeVThus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to a 92238U nucleus to make it emit a proton.

We are given the following atomic masses: 92238U=238.05079 u , 24He=4.00260 u 90234Th=234.04363 u , 11H=1.00783 u 91237Pa=237.05121 u. Take 1 u≡931.5 MeVHere the symbol Pa is for the element protactinium (Z = 91). Mark the CORRECT statement(s)

The energy released in α-decay is given by Q=(Mu−MTh−MHe)c2Substituting the atomic masses as given in the data, we findQ=(238.05079−234.04363−4.00260)u×c2=(0.00456 u)c2=(0.00456 u)c2 =(0.00456 u)(931.5 MeV/u)=4.25 MeV .If 92238U spontaneously emits a proton, the decay process would be 92238U→ 91237Pa+ 11HThe Q for this process to happen is =(MU−MPa−MH)c2=(238.05079−237.05121−1.00783)u×c2=(−0.00825 u)c2 =−(0.00825 u)(931.5 MeV/u) =−7.68 MeVThus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to a 92238U nucleus to make it emit a proton.

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We are given the following atomic masses:
$_{92}^{238} U = 238.05079 u,_{2}^{4} H e = 4.00260 u$
$_{90}^{234} T h = 234.04363 u,_{1}^{1} H = 1.00783 u$
$_{91}^{237} P a = 237.05121 u$ . Take $1 u \equiv 931.5 M e V$
Here the symbol Pa is for the element protactinium (Z = 91). Mark the CORRECT statement(s)

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The energy released during the $α$ -decay of $_{92}^{238} U$ is 4.75 MeV rounded to two places after decimal.

The energy released during the $α$ -decay of $_{92}^{238} U$ is 4.25 MeV rounded to two places after decimal.

The emission of proton from $_{92}^{238} U$ can be spontaneous.

The emission of proton from $_{92}^{238} U$ cannot be spontaneous.

answer is B.

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Detailed Solution

The energy released in $α$ -decay is given by $Q = (M_{u} - M_{T h} - M_{H e}) c^{2}$
Substituting the atomic masses as given in the data, we find
$Q = (238.05079 - 234.04363 - 4.00260) u \times c^{2} = (0.00456 u) c^{2} = (0.00456 u) c^{2}$
$= (0.00456 u) (931.5 M e V / u) = 4.25 M e V$ .

If $_{92}^{238} U$ spontaneously emits a proton, the decay process would be $_{92}^{238} U \to_{91}^{237} P a +_{1}^{1} H$
The Q for this process to happen is $= (M_{U} - M_{P a} - M_{H}) c^{2} = (238.05079 - 237.05121 - 1.00783) u \times c^{2} = (- 0.00825 u) c^{2}$
$= - (0.00825 u) (931.5 M e V / u) = - 7.68 M e V$

Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to a $_{92}^{238} U$ nucleus to make it emit a proton.

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