The energy that should be added to an electron to reduce its de Broglie wavelength from 1 nm to 0.5 nm is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → thrice the initial energy
According to De-Broglie wavelength,
$λ=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$
After decreasing wavelength,
$λ'=\frac{h}{p'}=\frac{h}{\sqrt{2mE'}}$
$∴\frac{λ}{λ'}=\sqrt{\frac{E'}{E}}$
$⇒\frac{E'}{E}=\left(\frac{λ}{λ'}\right)^2$
$⇒\frac{E'}{E}=\left(\frac{1}{0.5}\right)^2$
$⇒\frac{E'}{E}=4$
$⇒E'=4E$
∴ energy which should be added is,
$E=E'-E$
$=4E-E$
$=3E$