A) \[\frac{{{E}_{e}}}{{{E}_{ph}}}\,\,=\,\,\frac{2c}{v}\]
B) \[\frac{{{E}_{e}}}{{{E}_{ph}}}\,\,=\,\,\frac{v}{2c}\]
C) \[\frac{{{P}_{e}}}{{{P}_{ph}}}\,\,=\,\,\frac{2c}{v}\]
D) \[\frac{{{P}_{e}}}{{{P}_{ph}}}\,\,=\,\,\frac{v}{2c}\]
Correct Answer: B
Solution :
\[\frac{{{E}_{e}}}{{{E}_{p}}}=\frac{\frac{1}{2}m{{v}^{2}}}{h\upsilon }=\frac{1}{2}v\times \frac{mv}{h}.\frac{h}{h\upsilon }\] But \[\frac{h}{mv}\,\,=\,\,\frac{h}{h\upsilon }\] \[\therefore \,\,\frac{{{E}_{e}}}{{{E}_{ph}}}=\frac{1}{2}v\,\left[ \frac{mv}{h}.\frac{h}{h\upsilon }.\frac{c}{c} \right]=\frac{v}{2c}\]You need to login to perform this action.
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