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  •  An electron accelerated through a potential difference  has a de-Broglie wavelength of <img alt="\

 An electron accelerated through a potential difference V_{1} has a de-Broglie wavelength of \lambda. When the potential is changed to V_{2},  its de-Broglie wavelength increases by 50 \%.The value of \left(\frac{V_{1}}{V_{2}}\right) is equal to

Option: 1

3


Option: 2

\frac{3}{2}


Option: 3

4


Option: 4

\frac{9}{4}


Answers (1)

best_answer

\mathrm{KE}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}
\mathrm{P}=\frac{\mathrm{h}}{\lambda}
\mathrm{eV}_{1}=\frac{\left(\frac{\mathrm{h}}{\lambda}\right)^{2}}{2 \mathrm{~m}}
\mathrm{eV}=\frac{\left(\frac{\mathrm{h}}{1.5 \lambda}\right)^{2}}{2 \mathrm{~m}}
\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=(1.5)^{2}=\frac{9}{4}

 

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Rishi

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