# NEET Questions Solved

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

(a) Four times the initial energy

(b) Equal to the initial energy

(c) Twice the initial energy

(d) Thrice the initial energy

(d) $\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}};\frac{\mathrm{\lambda }\text{'}}{\mathrm{\lambda }}=\sqrt{\frac{\mathrm{E}}{\mathrm{E}\text{'}}}⇒\frac{\mathrm{E}}{\mathrm{E}}={\left(\frac{0.5}{1}\right)}^{2}⇒\mathrm{E}\text{'}=\frac{\mathrm{E}}{0.25}=4\mathrm{E}$

The energy should be added to decrease wavelength.
= E' - E = 3E

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