Curium-242 undergoes alpha decay. In addition to the alpha particle, which
2 answers:
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Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the orbit of hydrogen atom.
Formula used :
where,
= energy of
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.
The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV