How many disintegrations per second will occur in one gram of ${}_{92}U^{238},$ if half life against alpha decay 8is 1.42 × 10¹⁷ s?

Solution

Given, half life period (T) = $\frac{0.693}{λ} = 1.42 \times 10^{17} s$

Therefore,

$λ = \frac{0.693}{1.42 \times 10^{17}} = 4.88 \times 10^{- 18}$

Avogadro's number = $6.023 \times 10^{23}$

n is the number of atoms present in 1 g of ${}_{92}^{238}U$ = $\frac{N}{A} = \frac{0.623 \times 10^{23}}{238} = 25.30 \times 10^{20}$

Number of disintegrations is given by,

$\frac{dN}{dt} = λ n = 4.88 \times 10^{- 18} \times 25.30 \times 10^{20} = 1.2346 \times 10^{4} d i \sin t e g r a t i o n s p e r s e c$

For Daily Free Study Material Join wiredfaculty

Wave Optics

How many disintegrations per second will occur in one gram of ${}_{92}U^{238},$ if half life against alpha decay 8is 1.42 × 10¹⁷ s?

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Wave Optics

How many disintegrations per second will occur in one gram of U23892, if half life against alpha decay 8is 1.42 × 1017 s?

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

How many disintegrations per second will occur in one gram of ${}_{92}U^{238},$ if half life against alpha decay 8is 1.42 × 10¹⁷ s?