Question

An electron is continuously accelerated in vacuum tube by applying potential difference. If its de Broglie wavelength is decreased by 1%, the change in the kinetic energy of the electron is nearly:

• A. Decreased by 1.0%
• B. Increased by 2.0%
• C. Increased by 1.0%
• D. Decreased by 2.0%

Answer 1

The correct option is B Increased by 2.0%

According to de Broglie equation,
$\lambda =\frac{h}{mv}=\frac{h}{P}$ --(eq.1)
and kinetic energy $K=\frac{P^2}{2m}$
or $P=\sqrt{2mK}$
putting in (eq.1) we get
$\lambda =\frac{h}{\sqrt{2mK}}$
so according to question,
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\sqrt{K_1}}{\sqrt{K_2}}$
$\frac{0.99{\lambda }_1}{{\lambda }_1}=\frac{\sqrt{K_1}}{\sqrt{K_2}}$
$\sqrt{K_1}=0.99\sqrt{K_2}$
or $K_2=1.02K_1$
so, % change in K.E $=\frac{K_2-K_1}{K_1}\times 100=2\%$