In an experiment the initial number of radioactive nuclei is 3000- It is found that 1000- 40 nuclei decayed in the 1-0 s- For -x- - - 1- ln1-x - x up to first power in x- The error - in the determination of the decay constant - is s-1- is

The correct option is C 0.02 N = N_o e ^ λ t #10; ln N = lnN_o λ t #10; dNN = λ t #10;Converting to error #10; Δ NN = Δλ t ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Law of Radioactivity

In an experim...

Question

In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm 40$ nuclei decayed in the $1.0 s$ . For $| x | << 1, ln (1 + x) = x$ up to first power in $x$ . The error $Δ λ$ , in the determination of the decay constant $λ$ , is $s^{- 1}$ , is

0.04

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.03

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.02

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.01

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.02$
$N = N_{o} e^{- λ t}$ $l n N = l n N_{o} - λ t$ $\frac{d N}{N} = - λ t$ Converting to error $\frac{Δ N}{N} = Δ λ t$ $Δ λ = \frac{40}{2000 \times L}$ =0.02

Suggest Corrections

Similar questions

Q. PARAGRAPH
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation

z = x / y

. If the errors in

x, y

and

z

are

Δ x, Δ y

and

Δ z

, respectively, then

z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) {(1 \pm \frac{Δ y}{y})}^{- 1}

.
The series expansion for

{(1 \pm \frac{Δ y}{y})}^{- 1}

, to first power in

Δ y / y

1 \mp (Δ y / y)

. The relative errors in independent variables are always added. So the error in

z

will be

Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y})

.
The above derivation makes the assumption that

Δ x / x ≪ 1, Δ y / y ≪ 1

. Therefore, the higher powers of these quantities are neglected.

In an experiment the initial number of radioactive nuclei is

3000

. It is found that

1000 \pm 40

nuclei decayed in the

1.0 s

. For

| x | << 1, ln (1 + x) = x

up to first power in

x

. The error

Δ λ

, in the determination of the decay constant

λ

, in

s^{- 1}

z = x / y

. If the errors in

x, y

and

z

are

Δ x, Δ y

and

Δ z

, respectively, then

z \pm Δ z = \frac{x \pm Δ x}{y \pm Δ y} = \frac{x}{y} (1 \pm \frac{Δ x}{x}) {(1 \pm \frac{Δ y}{y})}^{- 1}

.
The series expansion for

{(1 \pm \frac{Δ y}{y})}^{- 1}

, to first power in

Δ y / y

1 \mp (Δ y / y)

. The relative errors in independent variables are always added. So the error in

z

will be

Δ z = z (\frac{Δ x}{x} + \frac{Δ y}{y})

.
The above derivation makes the assumption that

Δ x / x ≪ 1, Δ y / y ≪ 1

. Therefore, the higher powers of these quantities are neglected.

In an experiment the initial number of radioactive nuclei is

3000

. It is found that

1000 \pm 40

nuclei decayed in the

1.0 s

. For

| x | << 1, ln (1 + x) = x

up to first power in

x

. The error

Δ λ

, in the determination of the decay constant

λ

, in

s^{- 1}

Q. Two radioactive materials X

_{1}

and X

_{2}

have decay constants 10

λ

and

λ

respectively. If finally they have the same number of nuclei, then the ratio of the number of nuclei of X

_{1}

to that of X

_{2}

will be 1/e after a time

Q. The rate at which a particular decay process occurs in a radioactive sample, is proportional to the number of radioactive nuclei present. If N is the number of radioactive nuclei present at some instant, the rate of change of N is

\frac{d N}{d t} = - λ N

Consider radioactive decay of A to B which may further decay either to X or to Y.

λ_{1}, λ_{2}

and

λ_{3}

are decay constant for A to B decay, B to X decay and B to Y decay respectively. If at

t = 0

number of nuclei of A,B,X and Y are

N_{0}, N_{0}

, zero and zero respectively and

N_{1}, N_{2}, N_{3}, N_{4}

are number of nuclei A,B,X and Y at any instant.

Rate of accumulation of at any instant will be

Q. Obtain an expression for the number of radioactive nuclei present at any instant in terms of the decay constant and initial number of nuclei.

Join BYJU'S Learning Program

In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000±40 nuclei decayed in the 1.0 s. For |x|<<1,ln(1+x)=x up to first power in x. The error Δλ, in the determination of the decay constant λ, is s−1, is

The correct option is C 0.02N=Noe−λt lnN=lnNo−λt dNN=−λt Converting to error ΔNN=Δλt Δλ=402000×L =0.02

In an experiment the initial number of radioactive nuclei is $3000$ . It is found that $1000 \pm 40$ nuclei decayed in the $1.0 s$ . For $| x | << 1, ln (1 + x) = x$ up to first power in $x$ . The error $Δ λ$ , in the determination of the decay constant $λ$ , is $s^{- 1}$ , is

The correct option is C $0.02$
$N = N_{o} e^{- λ t}$ $l n N = l n N_{o} - λ t$ $\frac{d N}{N} = - λ t$ Converting to error $\frac{Δ N}{N} = Δ λ t$ $Δ λ = \frac{40}{2000 \times L}$ =0.02