The figure given below shows a radioactive nucleus A composed of 84 protons and 128 neutrons kept in a thick lead walled container. The emitted particles pass through a magnetic field in a direction perpendicular to the plane of paper inwards as shown by X. The nucleus A emits a particle which deflects to the left and is transformed into nucleus B. Nucleus B further emits a particle which deflects towards right and transforms into nucleus C.

(a) Name the radiations emitted by nucleus A and B.

(b) Name the law used to identify the radiations.

(c) What is the composition of nucleus C?

(a) Nucleus A emits alpha (α) radiation and nucleus B emits beta (β) radiation.
As particles emitted by nucleus A deflect to left hence, they are alpha (α) particles and nucleus B emission deflects to the right hence, they are beta (β) particles.

(b) Fleming's left hand rule.

(c) Nucleus C will have 83 protons and 125 neutrons.

Explanation:

Given,

Before any decay :

Number of protons in A = 84 = Atomic number of A

Number of neutrons in A = 128

Mass number = 84 + 128 = 212

After alpha decay (A → B) :

Loss of protons = 2

Loss of neutrons = 2

New atomic number = 84 – 2 = 82

New mass number = 212 – 4 = 208

After beta decay (B → C) :

As in beta decay, a neutron converts into a proton so that Atomic number increases by 1 and mass number remains the same.

New atomic number = 82 + 1 = 83

Mass number = 208

For C

Number of protons = atomic number of C = 83

Number of neutrons = Mass number - number of proton = 208 - 83 = 125