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 ...
Yes, your calculation is correct! The half-life of a nuclide can be determined using the decay constant. The decay constant is the probability of decay per unit time and is usually denoted by the symbol λ (lambda).
To find the half-life, you can use the formula:
t1/2 = ln(2) / λ
Where t1/2 is the half-life and ln(2) is the natural logarithm of 2 (approximately 0.693).
In this case, you are given that the decay constant (λ) is 10^-6 s^-1. So you can calculate the half-life as:
t1/2 = ln(2) / 10^-6
t1/2 ≈ (0.693) / (10^-6)
t1/2 ≈ 6.93 x 10^5 seconds
So, your answer is correct! The half-life of the nuclide with a decay constant of 10^-6 s^-1 is approximately 6.93 x 10^5 seconds.