The kinetic energy of an electron, α - particle and a proton are given as 4 K ,   2 K and K respectively. The de-Broglie wavelength associated with electron ( λ e ) ,   α - particle ( λ α ) and the proton ( λ p ) are as follows:

Question

The kinetic energy of an electron, &#945; - particle and a proton are given as 4 K , &#160; 2 K and K respectively. The de-Broglie wavelength associated with electron ( &#955; e ) , &#160; &#945; - particle ( &#955; &#945; ) and the proton ( &#955; p ) are as follows:

Aicharya · Accepted Answer

The formula to calculate the de Broglie wavelength of a particle is given by
$λ = \frac{h}{\sqrt{2 m K}} . . . (1)$
Now, the masses of the given particles are interrelated by the following relations:
$\begin{array}{rcl} m_{e} & \approx & \frac{m_{p}}{1837} \\ m_{α} & = & 4 m_{p} \end{array}$
The de Broglie wavelength for electron is given by
$\begin{array}{rcl} λ_{e} & = & \frac{h}{\sqrt{2 m_{e} (4 K)}} \\ = & \frac{h}{\sqrt{\frac{8}{1837} m_{p} K}} . . . (2) \end{array}$
The de Broglie wavelength for proton is given by
$\begin{array}{rcl} λ_{p} & = & \frac{h}{\sqrt{2 m_{p} (2 K)}} \\ = & \frac{h}{\sqrt{4 m_{p} K}} . . . (3) \end{array}$
And, the de Broglie wavelength of alpha particle is given by
$\begin{array}{rcl} λ_{α} & = & \frac{h}{\sqrt{2 m_{α} K}} \\ = & \frac{h}{\sqrt{2 (4 m_{p}) K}} \\ = & \frac{h}{\sqrt{8 m_{p} K}} . . . (4) \end{array}$
From equation (2), (3) and (4), it can be concluded that $λ_{e} > λ_{p} > λ_{α}$ .

The kinetic energy of an electron, α - particle and a proton are given as 4 K ,   2 K and K…

Solution

The kinetic energy of an electron, &#945; - particle and a proton are given as 4 K , &#160; 2 K and K…

Solution

The kinetic energy of an electron, α - particle and a proton are given as 4 K , 2 K and K…