Initial amount, [tex]A_o=150\ g[/tex] .
Final amount, [tex]A =10\ g[/tex] .
Time taken, t = 6:00 - 2:00 = 4 hour.
We know,
[tex]A=A_o(\dfrac{1}{2})^{\dfrac{t}{h}}\\\\10 = 150 \times \dfrac{1}{2}^{\dfrac{4}{h}}\\\\2^{\dfrac{4}{h}}=15\\\\\dfrac{4}{h}= log_215\\\\h = \dfrac{4}{log_215}\\\\h = 1.024 \ hours[/tex]
Therefore, the half-life of the isotope is 1.024 hours.
Hence, this is the required solution.
