If a nuclide has a decay constant of 10^-6 s-1 what is its half life ???
I think it is 0.693 / 10^-6 = 693000 or 6.93 x 10^5
Is this right ...
To calculate the half-life of a nuclide given its decay constant, you can use the formula:
T1/2 = ln(2) / λ
where T1/2 is the half-life, ln is the natural logarithm, and λ is the decay constant.
In this case, the decay constant is given as 10^-6 s^-1. Plugging this value into the formula:
T1/2 = ln(2) / (10^-6)
To simplify this, ln(2) is approximately 0.693.
T1/2 = 0.693 / (10^-6)
To divide by a decimal in scientific notation, you can multiply by its reciprocal. In this case, the reciprocal of 10^-6 is 10^6.
T1/2 = 0.693 x 10^6
Converting this to scientific notation, the half-life is:
T1/2 = 6.93 x 10^5 seconds
So, your calculation is correct. The half-life of the nuclide with a decay constant of 10^-6 s^-1 is 6.93 x 10^5 seconds.