Question

Calculate the energy released if U238-nucleus emits an α−particle.

OR
Calculate the energy released in MeV in the following nuclear reaction
92238​U→90234​Th+24​He+Q
Given Atomic mass of 238U=238.05079u
Atomic mass of 234Th=234.04363u
Atomic mass of alpha particle =4.00260u
1u=931.5MeV/c2
Is the decay spontaneous ? Give reasons.

Answer 1

The energy released in the decay of 238U to 234Th and an alpha (α) particle is approximately 0.0259 MeV.

Step-by-step explanation:

In the given nuclear reaction, 238U decays into 234Th and emits an alpha (α) particle, 4He.

To calculate the energy released in this decay, we first need to find the change in mass (Am) of the system. The atomic mass of 238U is 238.05079u, the atomic mass of 234Th is 234.04363u, and the atomic mass of the alpha particle is 4.00260u.

The energy released (Q) can be calculated using the equation Q = Am × c², where c is the speed of light and is approximately 3 × 10^8 m/s.

Using the given atomic masses, we find that Am = (238.05079u - 234.04363u - 4.00260u) = 0.00456u. Converting this to kilograms and using the equation E = mc², we find the energy released to be approximately 4.12 × 10^-11 Joules or 0.0259 MeV.