There are two radio nuceli `A` and `B. A` is an `alpha` emitter and `B` a `beta` emitter. Their disintegration constant are in the ratio of `1:2` What should be the ratio of number of atoms of `A` and `B` at any time t so that probabilities of getting alpha and beta particles are same at that instant?

There are two radio nuclei A and B is a alpha -emitter and B is beta -emitter. Their disintegration constant are in the ratio of 1:2 . What should be the number of atoms ratio of A and B at time t=0 , so that initially probability of getting of alpha and beta -particles are same.

There are two radio-active nuclei A and B . A in an alfa-emitter while B is a beta-emitter, Their disintegration constant are in the ratio of 1:4 . The ratio of number of nuclei of A and B at any time 't' such that probabilities of getting number of alpha and beta particles are same at the instant is y:1 . Then y ______is

Two Radioactive substances X and Y emit a and B particles respectively. Their disintegration constants are in the ratio 2 : 3. To have equal rate of disintegration of getting emission of a and particles, the ratio of number of atoms of X to that of Y at any time instant is

Ionized hydrogen atoms and alpha -particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths r_(H):r_(alpha) will be :

A nuclide A (with mass number m and atomic number n ) disintegrates emitting an alpha and one beta particle. The resulting nuclide B has mass number and atomic number respectively equal to