Question

A proton and an electron are both accelerated to the same final speed. If λp s the de Broglie wavelength of the proton and e is the de Broglie wavelength of the electron, then 5, D) Not enough data to answer this question

Answer 1

Answer:

Explanation:

The relation between the de Broglie wavelength and the momentum of the particle is given by

$\lambda =\frac{h}{m\times v}$

where, m is the mas of the particle and v be the velocity of the particle and h be the Plank's constant.  

So, the de broglie wavelength of proton is given by

$\lambda _{p}=\frac{h}{m_{p}\times v}$ .... (1)

The de broglie wavelength of electron is given by

$\lambda _{e}=\frac{h}{m_{e}\times v}$ .... (2)

Divide equation (2) by equation (1), we get

$\frac{\lambda _{e}}{\lambda _{p}}=\frac{m_{p}}{m_{e}}$

As the mass of proton is much more than the mass of electron, so the de broglie wavelength of electron is more than the de Broglie wavelength of proton.