Question

If the energy of a particle is reduced to one fourth, then the percentage increase in its de Broglie wavelength will be :

• A. $4$1%
• B. $141$%
• C. $100$%
• D. $71$%

Answer 1

The correct option is C $100$%

We know,

$\frac{1}{2^{\cdot }}m{v}^2=E$

$mv=\sqrt{2mE}$ &

$\lambda =\frac{h}{mv}$

${\lambda }^{\prime }=\frac{h}{\sqrt{2m{E}^{\prime }}}$

${\lambda }^{{}^{\prime }}=\frac{h}{\sqrt{2m\frac{E}{4}}}$

$\frac{{\lambda }^{{}^{\prime }}}{\lambda }=2$

${\lambda }^{{}^{\prime }}=2\lambda$

So, the answer is option (C).