Question

Half life of F is 110 minutes. What
fraction of IF sample decays in 20
minutes ?
(Ans. : 0.118)​

Answer 1

half life of F is 110 minutes.

using formula of time period , T = ln2/K

⇒110 = ln2/k

⇒k = ln2/110 min^-1

hence, decay constant , k = 0.693/110 = 6.3 × 10^-3 min^-1

t = 2.303/k log[1/fraction ]

⇒20 min= 2.303/6.3 × 10^-3 log[1/fraction]

⇒ 20 × 6.3 × 10^-3/2.303 = log[1/fraction]

⇒54.711 × 10^-3 = log[1/fraction ]

⇒1/fraction = 10^(54.711 × 10^-3)

⇒fraction = 10^(-54.711 × 10^-3)

⇒fraction ≈ 0.8816

hence, 0.8816F is remaining after 20 min .

then, fraction of F sample decays in 20 min = 1 - 0.8816 = 0.1184 ≈ 0.118 [ Ans]

Answer 2

$\huge\bold\purple{Answer:-}$

half life of F is 110 minutes.

using formula of time period , T = ln2/K

⇒110 = ln2/k

⇒k = ln2/110 min^-1

hence, decay constant , k = 0.693/110 = 6.3 × 10^-3 min^-1

t = 2.303/k log[1/fraction ]

⇒20 min= 2.303/6.3 × 10^-3 log[1/fraction]

⇒ 20 × 6.3 × 10^-3/2.303 = log[1/fraction]

⇒54.711 × 10^-3 = log[1/fraction ]

⇒1/fraction = 10^(54.711 × 10^-3)

⇒fraction = 10^(-54.711 × 10^-3)

⇒fraction ≈ 0.8816