Analyzing the \( \rm ^{87}_{36} {Kr} \rightarrow^{86}_{36} Kr \) Nuclear Decay

The question asks us to identify the type of nuclear decay process when the nuclide \( \rm ^{87}_{36} {Kr} \) transforms into \( \rm ^{86}_{36} Kr \). Nuclear decay involves changes in the composition of an atomic nucleus.

The notation \( \rm ^{A}_{Z} X \) represents a nuclide, where:

• A is the mass number (total number of protons and neutrons).
• Z is the atomic number (number of protons).
• X is the chemical symbol of the element.

Let's examine the initial and final nuclides in the given transformation:

• Initial nuclide: \( \rm ^{87}_{36} {Kr} \)
  • Mass number (\(A_i\)) = 87
  • Atomic number (\(Z_i\)) = 36
• Final nuclide: \( \rm ^{86}_{36} Kr \)
  • Mass number (\(A_f\)) = 86
  • Atomic number (\(Z_f\)) = 36

Now, let's determine the changes in the mass number (\(\Delta A\)) and the atomic number (\(\Delta Z\)):

• Change in Mass Number: \( \Delta A = A_f - A_i = 86 - 87 = -1 \)
• Change in Atomic Number: \( \Delta Z = Z_f - Z_i = 36 - 36 = 0 \)

So, the transformation involves a decrease of 1 in the mass number and no change in the atomic number.

Comparing Observed Changes with Different Nuclear Decay Types

Let's look at how the mass number (\(A\)) and atomic number (\(Z\)) change in common types of nuclear decay:

• α-decay: A nucleus emits an alpha particle (\( \rm ^{4}_{2} He \)).
  • \(\Delta A = -4\)
  • \(\Delta Z = -2\)

  \(\rm ^{A}_{Z} X \rightarrow^{A-4}_{Z-2} Y + ^{4}_{2} He\)

  This does not match \(\Delta A = -1\) and \(\Delta Z = 0\).

• β-decay (β\(^-\)): A neutron transforms into a proton, emitting an electron (\( \rm ^{0}_{-1} e \)) and an antineutrino.
  • \(\Delta A = 0\)
  • \(\Delta Z = +1\)

  \(\rm ^{A}_{Z} X \rightarrow^{A}_{Z+1} Y + ^{0}_{-1} e + \bar{\nu}_e\)

  This does not match \(\Delta A = -1\) and \(\Delta Z = 0\).

• Positron emission (β\(^+\)): A proton transforms into a neutron, emitting a positron (\( \rm ^{0}_{+1} e \)) and a neutrino.
  • \(\Delta A = 0\)
  • \(\Delta Z = -1\)

  \(\rm ^{A}_{Z} X \rightarrow^{A}_{Z-1} Y + ^{0}_{+1} e + \nu_e\)

  This does not match \(\Delta A = -1\) and \(\Delta Z = 0\).

• Electron Capture: A nucleus captures an inner atomic electron, a proton combines with the electron to form a neutron, and a neutrino is emitted.
  • \(\Delta A = 0\)
  • \(\Delta Z = -1\)

  \(\rm ^{A}_{Z} X + e^- \rightarrow^{A}_{Z-1} Y + \nu_e\)

  This does not match \(\Delta A = -1\) and \(\Delta Z = 0\).

• Neutron-emission: A nucleus emits a neutron (\( \rm ^{1}_{0} n \)).
  • \(\Delta A = -1\) (since the mass number of a neutron is 1)
  • \(\Delta Z = 0\) (since the atomic number of a neutron is 0)

  \(\rm ^{A}_{Z} X \rightarrow^{A-1}_{Z} X + ^{1}_{0} n\)

  This matches the observed changes: \(\Delta A = -1\) and \(\Delta Z = 0\).

Conclusion on Kr-87 to Kr-86 Transformation

The transformation of \( \rm ^{87}_{36} {Kr} \) to \( \rm ^{86}_{36} Kr \), which involves a decrease of 1 in mass number and no change in atomic number, is consistent with the emission of a neutron. The nuclear equation for this process is:

\(\rm ^{87}_{36} {Kr} \rightarrow^{86}_{36} Kr + ^{1}_{0} n\)

This confirms that the decay type is neutron-emission.