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Standard XII

Chemistry

Introduction to Quantum Mechanical Model

Gold 79198A...

Question

Gold $_{79}^{198} A u$ undergoes $β^{-}$ decay to an excited state of $_{80}^{198} H g$ . If the excited state, decays by emission of a $γ$ - photon with energy $0.412 M e V,$ the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligible energy. The recoil energy of the $_{80}^{198} H g$ nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968225 u for $_{79}^{198} A u$ and $197.966752 u$ for $_{79}^{198} H g$ .)

0.412 M e V

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1.371 M e V

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0.961 M e V

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1.473 M e V

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Solution

The correct option is C $0.961 M e V$
Given reaction is:
$_{79}^{198} A u ⟶_{80}^{198} H g +_{- 1}^{0} e + γ$
Energy obtained is, $(M_{_{79}^{198} A u} - M_{_{80}^{198} H g}) c^{2}$ $= (197.968225 u - 197.966752 u) {(3 \times 10^{8})}^{2}$
And, $1 u = 1.6605 \times 10^{- 27} k g$
Therefore, energy obtained in reduction of mass is $2.2 \times 10^{- 13} J$ , which is $1.373 M e V$
Photon carries $0.412 M e V$ , which leaves electron with energy $1.373 - 0.412 M e V$
Hence, electron has energy of $0.961 M e V$ .

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Similar questions

The isotope $_{5}^{12} B$ having a mass $12.014 u$ undergoes $β -$ decay to $_{6}^{12} C ._{6}^{12} C$ has an excited state of the nucleus $(_{6}^{12} C^{*})$ at $4.041 MeV$ above its ground state. If $_{5}^{12} B$ decay to $_{6}^{12} C^{*},$ the maximum kinetic energy of the $β -$ particle in units of $MeV$ nearly is (integer only).

Take : $m (_{6}^{12} C) = 12.0 u (1 u = 931.5 MeV / c^{2}, Where c is the speed of light in vacuum)$

Q. The isotope

_{5}^{12} B

having a mass 12.014 u undergoes

β

-decay to

_{6}^{12} C

_{6}^{12} C

has an excited state of the nucleus (

_{5}^{12} C^{*}

) at

4.041 M e V

above its ground state. If

_{5}^{12} B

decays to

_{6}^{12} C^{*}

, the maximum kinetic energy of the

β

-particle in units of

M e V

is
(

1 u = 931.5 M e V / c^{2}

, where c is the speed of light in vacuum).

Q. The isotope

_{5}^{12} B

having a mass 12.014 u undergoes

β

-decay to

_{6}^{12} C

_{6}^{12} C

has an excited state of the nucleus (

_{5}^{12} C^{*}

) at 4.041 MeV above its ground state. If

_{5}^{12} B

decays to

_{6}^{12} C^{*}

, the maximum kinetic energy of the

β

-particle in units of MeV is
(1 u = 931.5 MeV/

c^{2}

, where c is the speed of light in vacuum).

Suppose a $_{88}^{226} R a$ nucleus at rest and in ground state undergoes $α$ -decay to a $_{86}^{222} R n$ nucleus in its excited state. The kinetic energy of the emitted $α$ particle is found to be $4.44 M e V$ . $_{86}^{222} R n$ nucleus then goes to its ground state by $γ$ -decay. The energy of the emitted $γ$ photon is $k e V$ .

[Given : atomic mass of $_{88}^{226} R a = 226.005 u$ , atomic mass of $_{86}^{222} R n = 222.000 u$ , atomic mass of $α$ particle = $4.000 u, 1 u = 931 M e V / c^{2}, c$ is speed of light]

Q. The nucleus

^{23} N e

decays by

β -

emission into the nucleus

^{23} N a

. Write down the

β -

decay equation and determine the maximum kinetic energy of the electrons emitted. Given,

m (_{10}^{23} N e) = 22.994466 a m u

and

m (_{11}^{23} N a) = 22.989770 a m u

. Ignore the mass of antineutrino

(¯ ¯ ¯ v)

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