Answer:
[tex]\lambda' =1.21\times 10^{-10}\ m [/tex]
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,
[tex]P = \dfrac{h}{\lambda}[/tex]
Kinetic energy
[tex]K =\dfrac{p^2}{2m}[/tex]
[tex]K =\dfrac{h^2}{2m\lambda^2}[/tex]...........(1)
Kinetic energy of other particle
[tex]3K =\dfrac{h^2}{2m\lambda'^2}[/tex]............(2)
dividing equation (1)/(2)
[tex]\dfrac{1}{3} =\dfrac{\lambda'^2}{\lambda^2}[/tex]
[tex]\lambda' = \dfrac{\lambda'}{\sqrt{3}}[/tex]
[tex]\lambda' = \dfrac{2.1\times 10^{-10}}{\sqrt{3}}[/tex]
[tex]\lambda' =1.21\times 10^{-10}\ m [/tex]
