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  • The de-Broglie wavelength of a paraticle having kinetic energy E is . How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75%

The de-Broglie wavelength of a paraticle having kinetic energy E is \lambda. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the intial value?
Option: 1 E
Option: 2 \frac{7}{9}E
Option: 3 \frac{16}{9}E
Option: 4 \frac{1}{9}E

Answers (1)

best_answer

De-Broglie\: wavelength\: \left ( \lambda \right )= \frac{h}{\sqrt{2mE}}
\lambda_{i}= \frac{h}{\sqrt{2mE}}\rightarrow \left ( 1 \right )
\lambda_{f}=0\cdot 75\lambda_{i}= \frac{h}{\sqrt{2mE_{f}}}\rightarrow \left ( 2 \right )
\frac{1}{0\cdot 75}= \frac{\sqrt{E_{f}}}{\sqrt{E}}
\left ( \frac{4}{3} \right )^{2}E= E_{f}
E_{f}= \frac{16}{9}E
Extra energy is given = E_{f}= E_{i}
                              = \frac{16}{9}E-E
                              = \frac{7E}{9}
The correct option is (2)  

Posted by

vishal kumar

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