The energy that should be added to an electro | vedclass

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(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.From this relatio

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Thrice the initial energy

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(D) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{\sqrt{2mE}}$,where $E$ is the kinetic energy of the electron.From this relation,we have $\lambda \propto \frac{1}{\sqrt{E}}$,which implies $\frac{\lambda'}{\lambda} = \sqrt{\frac{E}{E'}}$.Given $\lambda = 1 \ nm$ and $\lambda' = 0.5 \ nm$,we substitute these values:$\frac{0.5}{1} = \sqrt{\frac{E}{E'}}$Squaring both sides,we get $\frac{1}{4} = \frac{E}{E'}$,which means $E' = 4E$.The energy that should be added is $\Delta E = E' - E = 4E - E = 3E$.Therefore,the energy to be added is thrice the initial energy.

The energy that should be added to an electron to reduce its de Broglie wavelength from $1 \ nm$ to $0.5 \ nm$ is

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