${}^{238}U$ has $92$ protons and $238$ nucleons. It decays by emitting an Alpha particle and becomes
${}^{238}U$ has $92$ protons and $238$ nucleons. It decays by emitting an Alpha particle and becomes
- [AIIMS 2006]
- A
${}_{92}^{234}U$
- B
${}_{90}^{234}Th$
- C
${}_{92}^{235}U$
- D
${}_{93}^{237}Np$
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]
In the nuclear reaction: $X(n,\,\alpha ){\,_3}L{i^7}$ the term $X$ will be
In the nuclear reaction: $X(n,\,\alpha ){\,_3}L{i^7}$ the term $X$ will be
- [AIEEE 2005]
A plot of the number of neutrons $(N)$ against the number of protons ( $P$ )of stable nuclei exhibits upward deviation from linearity for atomic number, $Z>20$. For an unstable nucleus having $N / P$ ratio less than $1$ , the possible mode($s$) of decay is(are)
($A$) $\beta^{-}$-decay ( $\beta$ emission)
($B$) orbital or $K$-electron sasture
($C$) neutron emission
($D$) $\beta^{+}$-decay (positron emission)
A plot of the number of neutrons $(N)$ against the number of protons ( $P$ )of stable nuclei exhibits upward deviation from linearity for atomic number, $Z>20$. For an unstable nucleus having $N / P$ ratio less than $1$ , the possible mode($s$) of decay is(are)
($A$) $\beta^{-}$-decay ( $\beta$ emission)
($B$) orbital or $K$-electron sasture
($C$) neutron emission
($D$) $\beta^{+}$-decay (positron emission)
- [IIT 2016]
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
Pauli suggested the emission of nutrino during $\beta^{+}$decay to explain
Pauli suggested the emission of nutrino during $\beta^{+}$decay to explain