JEE MAIN - Physics (2021 - 26th August Evening Shift - No. 2)
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?
$${1 \over 9}$$E
$${7 \over 9}$$E
E
$${16 \over 9}$$E
Explanation
$$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$, $$mv = \sqrt {2mE} $$
$$\lambda \propto {1 \over {\sqrt E }}$$
$${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}$$, $${\lambda _2} = 0.75{\lambda _1}$$
$${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}$$
$${E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E$$ (E1 = E)
Extra energy given = $${{16} \over 9}E - E = {7 \over 9}E$$
$$\lambda \propto {1 \over {\sqrt E }}$$
$${{{\lambda _2}} \over {{\lambda _1}}} = \sqrt {{{{E_1}} \over {{E_2}}}} = {3 \over 4}$$, $${\lambda _2} = 0.75{\lambda _1}$$
$${{{E_1}} \over {{E_2}}} = {\left( {{3 \over 4}} \right)^2}$$
$${E_2} = {{16} \over 9}{E_1} = {{16} \over 9}E$$ (E1 = E)
Extra energy given = $${{16} \over 9}E - E = {7 \over 9}E$$
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