For a radioactive element, a graph of $log N$ against time has a slope equal to:

+ 2.303 λ

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

+ \frac{λ}{2.303}

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

- \frac{λ}{2.303}

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

- 2.303 λ

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

Open in App

Solution

The correct option is D $- \frac{λ}{2.303}$
$log N_{0} - log N = \frac{λ t}{2.303}$
$log N = (\frac{- λ}{2.303}) t + log N_{0}$
$Y = M x + C$
$∴$ Slope $(M) = \frac{- λ}{2.303}$

Suggest Corrections

Similar questions

Q. For a first order reaction, the plot of ‘t’ against log C gives a straight line with slope equal to:

Q. Radioactivity of a radioactive element remains

1 / 10

of the original radioactivity after

2.303 s

. The half life period of radioactive element in second is:

Q. Radioactivity of a radioactive element remains

1 / 10

of the original radioactivity after

2.303

seconds. The half life period is

Q. Suppose

f (n) = {log}_{2} (3) . {log}_{3} (4) . {log}_{4} (5) . . . . . . . {log}_{n - 1} (n)

then the sum

\sum_{k = 2}^{100} f (2^{k}) = λ

then

\frac{λ}{1683}

is equal to?

Nuclei of a radioactive element are being produced at a constant rate $α$ . The element has a decay constant $λ$ . At time t = 0, there are $N_{0}$ nuclei of the element.

The number N of nuclei at time t is

Join BYJU'S Learning Program

For a radioactive element, a graph of logN against time has a slope equal to:

The correct option is D −λ2.303logN0−logN=λt2.303logN=(−λ2.303)t+logN0Y=Mx+C∴ Slope (M)=−λ2.303

For a radioactive element, a graph of $log N$ against time has a slope equal to:

The correct option is D $- \frac{λ}{2.303}$
$log N_{0} - log N = \frac{λ t}{2.303}$
$log N = (\frac{- λ}{2.303}) t + log N_{0}$
$Y = M x + C$
$∴$ Slope $(M) = \frac{- λ}{2.303}$