If travelling at equal speeds,the longest wav | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = h / p = h / (mv)$,where $h$ is Planck's constant,$m$ is the m

Vedclass · Accepted Answer

electron

Vedclass · Answer

(A) According to the de Broglie relation,the wavelength $\lambda$ is given by $\lambda = h / p = h / (mv)$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity.Since the speed $v$ is constant for all particles,the wavelength $\lambda$ is inversely proportional to the mass $m$ of the particle $(\lambda \propto 1/m)$.Comparing the masses: $m_{	ext{electron}} Since the electron has the smallest mass,it will have the longest wavelength.

If travelling at equal speeds,the longest wavelength of the following matter waves is that for a :

Similar Questions

What is the de-Broglie wavelength associated with the hydrogen electron in its third orbit?

What is the wavelength of an electron? What does it indicate?

The relationship between the momentum $(P)$ and the wavelength $(\lambda)$ of a particle is given by:

The de-Broglie's wavelength of an electron in the $4^{th}$ orbit is . . . . . . . . $\pi a_0$. ($a_0 =$ Bohr's radius)

The de Broglie wavelength of an electron with kinetic energy of $2.5 \ eV$ is (in $m$):
$(1 \ eV = 1.6 \times 10^{-19} \ J, m_{e} = 9 \times 10^{-31} \ kg)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module