Suppose a \({}_{88}^{226} \mathrm{Ra}\) nucleus at rest and in ground state undergoes \(\alpha\)-decay to a \({}_{86}^{222} \mathrm{Rn}\) nucleus in its excited state. The kinetic energy of the emitted \(\alpha\) particle is found to be 4.44 MeV. \({}_{86}^{222} \mathrm{Rn}\) nucleus then goes to its ground state by \(\gamma\)-decay. The energy of the emitted \(\gamma\)-photon is ____ keV, [Given: atomic mass of \({}_{88}^{226} \mathrm{Ra}=226.005 \mathrm{u}\), atomic mass of \({}_{86}^{222} \mathrm{Rn}=222.000 \mathrm{u}\), atomic mass of \(\alpha\) particle \(=4.000 \mathrm{u}\), \(1 \mathrm{u}=931 \mathrm{MeV} / \mathrm{c}^{2}\), c is speed of the light]

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Modern Physics

Suppose a

_{88}^{226} Ra

nucleus at rest and in ground state undergoes

α

-decay to a

_{86}^{222} Rn

nucleus in its excited state. The kinetic energy of the emitted

α

particle is found to be 4.44 MeV.

_{86}^{222} Rn

nucleus then goes to its ground state by

γ

-decay. The energy of the emitted

γ

-photon is ____ keV,
[Given: atomic mass of

_{88}^{226} Ra = 226.005 u

, atomic mass of

_{86}^{222} Rn = 222.000 u

, atomic mass of

α

particle

= 4.000 u

1 u = 931 MeV / c^{2}

, c is speed of the light]

Answer: 135

Q-value of the decay:

Q = (226.005 - 222.000 - 4.000) \times 931 = 0.005 \times 931 = 4.655 MeV

Kinetic energy of alpha particle:

K_{α} = \frac{A - 4}{A} (Q - E_{γ})

4.44 = \frac{222}{226} (Q - E_{γ})

Q - E_{γ} = 4.44 \times \frac{226}{222} = 4.520 MeV

E_{γ} = 4.655 - 4.520 = 0.135 MeV = 135 keV

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer