53 P 135 ⟶ Q + a ; t 1/2 = 1000 h rs Q ⟶ R + b ; t 1/2 = 10 min Calculate | TestHub

Passage / Comprehension

Radioactive elements can be categorized into:

(i) Neutron-rich nuclide

(ii) Neutron-poor nuclide

In order to achieve a stable n/p ratio, these nuclides either emit $β^{-}$ , $β^{+}$ (positron), or $α$ -particles, depending upon the n/p ratio of the unstable nuclei and stable nuclei.

₅₃P¹³⁵ $⟶$ Q + a

Q $⟶$ R + b

[Note: The only stable isotopes of elements P and Q are ₅₃P¹³³ and ₅₄Q¹³⁷, respectively.]

Elements P, Q, and R do not show resemblance to any known element.

$_{53} P^{135} ⟶ Q + a$ ; $t_{1/2} = 1000 h rs$

$Q ⟶ R + b$ ; $t_{1/2} = 10 min$

Calculate the number of nuclei of Q and R, respectively, after 1000 hours if we start with 2 mol of P.

[ $N_{A} = 6 \times 1 0^{23}$ ]

Options:

Answer:

Solution:

For 'Q' using steady state concept

Rate of appereance of $Q =$ Rate of disappeareance of Q

$K_{1} [P] = K_{2} [Q]$

Number of nuclei of $Q = \frac{k _{1}}{k _{2}} \times$ no. of nuclei of P

$= \frac{10/60}{1000} \times 6 \times 1 0^{23} = 1 0^{20}$

For ' R '

Since rate constant is very high for second step than first step

Number of nuclei of $R =$ Number of nuclei of P disintegrated $= 6 \times 1 0^{23}$

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

Doubts & Discussion