JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 16)
A particle moving with kinetic energy E has
de Broglie wavelength $$\lambda $$. If energy $$\Delta $$E is added
to its energy, the wavelength become $$\lambda $$/2. Value
of $$\Delta $$E, is :
E
3E
2E
4E
Explanation
$$\lambda = {h \over {\sqrt {2mE} }}$$
Also, $${h \over {\sqrt {2m\left( {E + \Delta E} \right)} }}$$ = $${\lambda \over 2}$$
$$ \therefore $$ $${{E + \Delta E} \over E} = 4$$
$$ \Rightarrow $$ $$\Delta $$E = 3E
Also, $${h \over {\sqrt {2m\left( {E + \Delta E} \right)} }}$$ = $${\lambda \over 2}$$
$$ \therefore $$ $${{E + \Delta E} \over E} = 4$$
$$ \Rightarrow $$ $$\Delta $$E = 3E
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