Question

if electron proton neutron and alpha particle are moving with same velocity then arrange the debroglie wavelength in increasing order

Answer 1

 m is mass, v is velocity, K is kinetic energy, p is momentum and λ is wavelength. Electrons have the smallest m, so they have the greatest λ. Alpha particles have the greatest m, so they have the smallest λ. ... Being as large as 43.4 times of proton and 86.8 times of alpha particle

Answer 2

Answer:

if electron proton neutron and alpha particle are moving with the same velocity then the de Broglie wavelength in increasing order is

$\ \lambda _e>\lambda _{\alpha }>\lambda _n=\lambda p\end{aligned}$

Explanation:

• The de Broglie wavelength is the wavelength of matter in wave nature

• For a particle with mass (m) and moving with velocity (v), the de Broglie wavelength is given by the formula

       λ =$h/p=h/mv$    (equation 1)

      where Planck's constant $h=6.6*10^{-34}$

     $p$- the momentum of the particle

From the question,

the velocity of both the particles is the same

but the masses are different

mass of electron $m_{e} =9.1*10^{-31} kg$

mass of neutron and proton are the same $m_{n}= m_{p}= 1.6*10^{-27} kg$

mass of the alpha particle $m_{\alpha } =2*9.1*10^{-31}kg$

since velocities are the same wavelength will depend only on the mass of the particle as they are inversely proportional to each other

$m_1>m_2\Rightarrow \lambda _1<\lambda _2$

$m_2>m_1\Rightarrow \lambda _2<\lambda _1$

that is maximum de Broglie wavelength is for a particle with less mass

decreasing order of mass and increasing order of wavelength is

$\begin{aligned}m_n=m_p<m_{\alpha }<m_e\\\Rightarrow \lambda _e>\lambda _{\alpha }>\lambda _n=\lambda p\end{aligned}$