CBSE-XI-Physics
47: The Special Theory of Relativity
- #16A particular particle created in a nuclear reactor leaves a 1 cm track before decaying. Assuming that the particle moved at 0.995c, calculate the life of the particle (a) in the lab frame and (b) in the frame of the particle. (a) in the lab frame and (b) in the frame of the particle.Ans : Given:
Length of the track, d = 1 cm
Velocity of the particle, v = 0.995c (a) Life of the particle in the lab frame is given by
`` t=\frac{d}{v}=\frac{0.01}{0.995c}``
`` =\frac{0.01}{0.995\times 3\times {10}^{8}}``
`` =33.5\times {10}^{-12}\,\mathrm{\,s\,}=33.5\,\mathrm{\,ps\,}`` (b) Let the life of the particle in the frame of the particle be t'. Thus,
`` t\text{'}=\frac{t}{\sqrt{1-{v}^{2}/{c}^{2}}}``
`` t\text{'}=\frac{33.5\times {10}^{-12}}{\sqrt{1-{\left(0.995\right)}^{2}}}``
`` t\text{'}=3.3541\times {10}^{-12}\,\mathrm{\,s\,}=3.3541\,\mathrm{\,ps\,}``
Page No 458: (a) Life of the particle in the lab frame is given by
`` t=\frac{d}{v}=\frac{0.01}{0.995c}``
`` =\frac{0.01}{0.995\times 3\times {10}^{8}}``
`` =33.5\times {10}^{-12}\,\mathrm{\,s\,}=33.5\,\mathrm{\,ps\,}`` (b) Let the life of the particle in the frame of the particle be t'. Thus,
`` t\text{'}=\frac{t}{\sqrt{1-{v}^{2}/{c}^{2}}}``
`` t\text{'}=\frac{33.5\times {10}^{-12}}{\sqrt{1-{\left(0.995\right)}^{2}}}``
`` t\text{'}=3.3541\times {10}^{-12}\,\mathrm{\,s\,}=3.3541\,\mathrm{\,ps\,}``
Page No 458:
- #16-ain the lab frame andAns : Life of the particle in the lab frame is given by
`` t=\frac{d}{v}=\frac{0.01}{0.995c}``
`` =\frac{0.01}{0.995\times 3\times {10}^{8}}``
`` =33.5\times {10}^{-12}\,\mathrm{\,s\,}=33.5\,\mathrm{\,ps\,}``
- #16-bin the frame of the particle.Ans : Let the life of the particle in the frame of the particle be t'. Thus,
`` t\text{'}=\frac{t}{\sqrt{1-{v}^{2}/{c}^{2}}}``
`` t\text{'}=\frac{33.5\times {10}^{-12}}{\sqrt{1-{\left(0.995\right)}^{2}}}``
`` t\text{'}=3.3541\times {10}^{-12}\,\mathrm{\,s\,}=3.3541\,\mathrm{\,ps\,}``
Page No 458: