A) \[\frac{1}{8}\]
B) \[\frac{3}{8}\]
C) \[\frac{5}{8}\]
D) \[\frac{7}{8}\]
Correct Answer: B
Solution :
Velocity of particle \[v=2.25\times {{10}^{8}}\] \[=\frac{3}{4}\times 3\times {{10}^{8}}=\frac{3}{4}c\] \[(\because \,c=3\times {{10}^{8}}\,m/s)\] de-Broglie wavelength of the particle is given by \[{{\lambda }_{1}}=\frac{h}{mv}\] ?(i) Energy of photon is given by as \[{{E}_{2}}=hv=\frac{hc}{{{\lambda }_{2}}}\] \[{{\lambda }_{2}}=\frac{hc}{{{E}_{2}}}\] ?(ii) but \[{{\lambda }_{1}}={{\lambda }_{2}}\] (given) \[\frac{h}{mv}=\frac{hc}{{{E}_{2}}}\] or \[{{E}_{2}}=mvc\] ?(iii) Now, kinetic energy of particle is \[{{E}_{1}}=\frac{1}{2}m{{v}^{2}}\] \[=\frac{1}{2}m{{\left( \frac{3}{4}c \right)}^{2}}\] \[{{E}_{1}}=\frac{9}{32}m{{c}^{2}}\] ?(iv) From Eqs. (iii) and (iv), we get \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{\frac{9}{32}m{{c}^{2}}}{mvc}=\frac{\frac{9}{32}m{{c}^{2}}}{m\times \frac{3}{4}c\times c}\] \[=\frac{9}{32}\times \frac{4}{3}=\frac{3}{8}\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{3}{8}\]
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