Question

A particle has a de Broglie wavelength of 2.1 × 10-10m. Then its kinetic energy increases by a factor of 3. What is the particle's new de Broglie wavelength, assuming that relativistic effects can be ignored?

Answer 1

Answer:

$\lambda' =1.21\times 10^{-10}\ m$

Explanation:

given,

de broglie wavelength of the particle = 2.1 × 10⁻¹⁰m

Kinetic energy increase factor = 3

de broglie  wavelength of new particle = ?

we know,

$P = \dfrac{h}{\lambda}$

Kinetic energy

$K =\dfrac{p^2}{2m}$

$K =\dfrac{h^2}{2m\lambda^2}$...........(1)

Kinetic energy of other particle

$3K =\dfrac{h^2}{2m\lambda'^2}$............(2)

dividing equation (1)/(2)

$\dfrac{1}{3} =\dfrac{\lambda'^2}{\lambda^2}$

$\lambda' = \dfrac{\lambda'}{\sqrt{3}}$

$\lambda' = \dfrac{2.1\times 10^{-10}}{\sqrt{3}}$

$\lambda' =1.21\times 10^{-10}\ m$