For a certain radioactive process the graph between In R and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately :

Solution

Physics sir · 4 steps

The slope of the ln R line is -λ, so read λ off the graph and divide 0.693 by it.

1. Why a straight line. The decay rate falls exponentially, R = R₀ e^(-λ t). Taking the natural log of both sides gives a straight line in t.

   ln R = ln R₀ - λ t

2. Read the slope. The line starts at ln R = 6 when t = 0 and reaches ln R = 0 at t = 40 s. So the slope is (0 - 6)/(40 - 0) = -0.15 s⁻¹, which means λ = 0.15 s⁻¹.
3. Half-life. T_(1/2) = (ln 2)/λ = 0.693/0.15 = 4.62 s.
4. So. The half-life is about 4.62 s, which is option (d).

The idea to keep. Any ln-versus-time graph for decay is a straight line whose slope magnitude is λ. Take the two intercepts, find the slope, and divide 0.693 by it.

Answer from the NTA final answer key. Solution by PrepWiser.