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  4. The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be:

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Q. The accelerating voltage of an electron is increased to two times, its de Broglie wavelength will be:

JIPMERJIPMER 2000

Solution:

From the relation according de Broglie law $ {{\lambda }_{d}}\propto \frac{1}{\sqrt{V}}...(i) $ [V becomes 2V (given)] $ \lambda {{}_{d}}\propto \frac{1}{\sqrt{2V}} $ ?(ii) $ \frac{{{\lambda }_{d}}}{\lambda {{}_{d}}}=\sqrt{\frac{2V}{V}} $ hence $ \lambda {{}_{d}}=\frac{{{\lambda }_{d}}}{\sqrt{2}} $ so $ \frac{\lambda {{}_{d}}}{{{\lambda }_{d}}}=\frac{1}{\sqrt{2}}=0.7\,\text{times} $