A nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is
A nucleus $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is
- [AIPMT 2011]
- A
$_n{X^{m - 4}}$
- B
$_{n - 2}{Y^{m - 4}}$
- C
$_{n - 4}{Z^{m - 4}}$
- D
$_{n - 1}{Z^{m - 4}}$
Similar Questions
A nucleus of an element ${}_{84}{X^{202}}$ emits an $\alpha $-particle first, $\beta $ -particle next and then a gamma photon. The final nucleus formed has an atomic number
A nucleus of an element ${}_{84}{X^{202}}$ emits an $\alpha $-particle first, $\beta $ -particle next and then a gamma photon. The final nucleus formed has an atomic number
In the given nuclear reaction, how many $\beta$ and $\alpha$ particles are emitted $_{92}{X^{235}} \to {\;_{82}}{Y^{207}}$
In the given nuclear reaction, how many $\beta$ and $\alpha$ particles are emitted $_{92}{X^{235}} \to {\;_{82}}{Y^{207}}$
When a uranium isotope ${ }_{92}^{235} U$ is bombarded with a neutron, it generates ${ }_{36}^{89} Kr$, three neutrons and:
When a uranium isotope ${ }_{92}^{235} U$ is bombarded with a neutron, it generates ${ }_{36}^{89} Kr$, three neutrons and:
- [NEET 2020]
Three $\alpha - $ particles and one $\beta - $ particle decaying takes place in series from an isotope $_{88}R{a^{238}}$. Finally the isotope obtained will be
Three $\alpha - $ particles and one $\beta - $ particle decaying takes place in series from an isotope $_{88}R{a^{238}}$. Finally the isotope obtained will be
A neutron strikes a ${ }_{92} U ^{235}$ nucleus and as a result ${ }_{38} Kr ^{93}$ and ${ }_{56} Ba ^{140}$ are produced with
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