Question

which of the following particle moving with same velocity has maximum de broglie wavelength​

Answer 1

Answer:

wavelength = h/mV

h & v are constant

Wavelength = ~ 1/m

jiska mass kam hoga uski wavelength jyada hogi

example = Electron

Answer 2

Beta particle has maximum de broglie wavelength

Explanation:

Suppose given particles are

(I). Electron

(II). Proton

(III). alpha particle

(IV). beta particle

If all particles are moving with same velocity

We need to calculated the maximum de broglie wavelength​

Using formula of wavelength

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

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

If all particles are moving with same velocity

Then, the wave length is inversely proportional to the mass of the particle

$\lambda\propto\dfrac{1}{m}$

We know that,

$m_{n}=m_{p}=1\ amu$

$\alpha = 4\ amu$

$\beta = \dfrac{1}{1200}\ amu$

The mass of the beta particle is very less so the beta particle has maximum de broglie wavelength.

Hence, Beta particle has maximum de broglie wavelength.

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Topic : de broglie

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