A radioactive element $X$ emits six $\alpha$-particles and four $\beta$-particles leading to ond product ${ }_{82}^{208} Pb$. $X$ is
A radioactive element $X$ emits six $\alpha$-particles and four $\beta$-particles leading to ond product ${ }_{82}^{208} Pb$. $X$ is
- A
${ }_{92}^{238} U$
- B
${ }_{90}^{230} Th$
- C
${ }_{90}^{232} Th$
- D
${ }_{92}^{239} U$
Similar Questions
How many alpha and beta particles are emitted when Uranium ${ }_{92} U ^{238}$ decays to lead ${ }_{82} Pb ^{206}$ ?
How many alpha and beta particles are emitted when Uranium ${ }_{92} U ^{238}$ decays to lead ${ }_{82} Pb ^{206}$ ?
- [JEE MAIN 2022]
A nucleus of an element $_{84}{X^{202}}$ emits an $\alpha-$ particle first, a $\beta-$ 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, a $\beta-$ particle next and then a gamma photon. The final nucleus formed has an atomic number
A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay $X_Z^A \rightarrow Y_{Z+1}^A+\beta^{-}+\bar{v}_e$ will occur is ( $m_e$ denotes the mass of the $\beta$ particle and the neutrino mass $m_v$ can be neglected)
A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay $X_Z^A \rightarrow Y_{Z+1}^A+\beta^{-}+\bar{v}_e$ will occur is ( $m_e$ denotes the mass of the $\beta$ particle and the neutrino mass $m_v$ can be neglected)
- [KVPY 2013]
Explain the basic nuclear process by equation in $\beta -$ decay.
Explain the basic nuclear process by equation in $\beta -$ decay.
$\gamma$-decay occurs when
$\gamma$-decay occurs when