KVPY Sample Paper KVPY Stream-SX Model Paper-4

  • question_answer
    A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time (in years) after which one-fourth of the material remains is

    A) 1080                

    B) 2430

    C) 3240                            

    D) 4860

    Correct Answer: A

    Solution :

    [a] \[\lambda ={{\lambda }_{1}}+{{\lambda }_{2}}\Rightarrow \frac{1}{t}=\frac{1}{{{t}_{1}}}+\frac{1}{{{t}_{2}}}\] \[\therefore t=\frac{{{t}_{1}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}=\frac{810\times 1620}{810+1620}=540\operatorname{year}\] Thus it takes two half-life to remains \[\frac{1}{4}th\]of the sample, so the time \[=2\times 540=1080\operatorname{year}\]


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