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  4. The additional energy that should be given to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is

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Q. The additional energy that should be given to an electron to reduce its de-Broglie wavelength from $1\, nm$ to $0.5\, nm$ is

KCETKCET 2013Dual Nature of Radiation and Matter

Solution:

The de-Broglie wavelength,
$\lambda=\frac{h}{\sqrt{2 m E_{k}}} \,\,\,\,\,\therefore \lambda \propto \frac{1}{\sqrt{E_{k}}}$
$(\because h$ and $m$ remains constant $)$
Here, in given condition
$\frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{E_{k_{2}}}{E_{k_{1}}}}, \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{4 E}{E}} $
$\lambda_{1}=2 \lambda_{2}, \lambda_{2}=\frac{\lambda_{1}}{2}$
So, if $E_{k}$ is increased by four times, then $\lambda$ becomes half.
$\therefore $ Additional kinetic energy that should be supplied to the electron
$=4 E_{k}-E_{k}=3 E_{k}$