When the radioactive `isotope _(88)Ra^(226)` decays in a series by emission of three aplha `(alpha)` and a beta `(beta)` particle, the isotope X which remains undecay is
A. `83^(X)^(214)`
B. `84^(X)^(218)`
C. `84^(X)^(220)`
D. `87^(X)^(223)`
Correct Answer - A
Due to emission os `3alpha` particles, the mass number is reduced by 12 while atomic number is decresed by 6. Due to emission of `beta` particle the atomic number is increased by one while there is no change in mass number is decreased by 5. so isotope X is is `X^(214)`
Correct Answer - B
`8 alpha` have been emitted
`4 beta^(-)` have been emitted
`2 beta^(+)` have emitted
`alpha` reduces atomic number by 2
`beta^(-)` increases atomic number by 1
`beta^(+)`...
The process for negative beta decay is given by equation
`._(19)^(40)K_(21) rarr ._(20)^(40)Ca_(20) +e^(-) + barv`
and the Q value is found from equation using atomic masses :
`Q_(beta) [m(.^(40)K)...
To determine the Q value for this decay, we first need to find the mass of the product nucleus `.^(12)C` in its excited state. In the ground state, `.^(12)C` has...
Correct Answer - A
Arrange the data as follows :
`alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5`
Median `=(1)/(2)` [value of 4th item+value of 5th item]
`therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`
Correct Answer - A
We have,
`A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]`
`=A(-alpha, -beta)`