Question 156: Dual Nature of Matter and Radiation

Question

The de-Broglie wavelength of a particle 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 initial value?

Accepted Answer

$$\lambda  = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$, $$mv = \sqrt {2mE} $$
$$\lambda  \propto {1 \over {\sqrt E }}$$
$${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}}  = {3 \over 4}$$, $${\lambda _2} = 0.75{\lambda _1}$$
$${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}$$
$${E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E$$ (E
1
= E)
Extra energy given = $${{16} \over 9}E - E = {7 \over 9}E$$

Answer

$${1 \over 9}$$E

Answer

$${7 \over 9}$$E

Answer

E

Answer

$${16 \over 9}$$E

Dual Nature of Matter and Radiation

The de-Broglie wavelength of a particle 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 initial value?

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