Respuesta :
Answer:
The wavelength of the particle is [tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]
Explanation:
We know that after accelerating across a potential 'V' the potential energy of the charge is converted into kinetic energy as
[tex]\frac{1}{2}mv^{2}=V\cdot e\\\\\therefore v=\sqrt{\frac{2V\cdot e}{m}}[/tex]
now according to De-Broglie theory the wavelength associated with a particle of mass 'm' moving with a speed of 'v' is given by
[tex]\lambda =\frac{h}{mv}[/tex]
where
'h' is planck's constant
'm' is the mass of particle
'v' is the velocity of the particle
Applying the values in the above equation we get
[tex]\lambda =\frac{h}{m\cdot \sqrt{\frac{2V\cdot e}{m}}}[/tex]
Thus
[tex]\lambda=\frac{h}{\sqrt{2V\cdot e\cdot m}}[/tex]
Answer:
Wavelength of the particle is [tex]\frac{h}{\sqrt{2meV}}[/tex]
Solution:
As per the question:
The particle with mass, m and charge, e accelerates through V (potential difference).
The momentum of the particle, [tex]p_{p}[/tex] if it travels with velocity, [tex]v_{p}[/tex]:
[tex]p_{p} = mv_{p}[/tex]
Now, squaring both sides and dividing by 2.
[tex]\frac{1}{2}p_{p}^{2} = \frac{1}{2}m^{2}v_{p}^{2}[/tex]
[tex]K.E =\frac{1}{2m}p_{p}^{2} = \frac{1}{2}mv_{p}^{2}[/tex]
[tex]K.E = \frac{1}{2}mv_{p}^{2}[/tex]
Also,
[tex]p_{p} = \sqrt{2mK.E}[/tex] (1)
Now, we know that Kinetic energy of particle accelerated through V:
K.E = eV (2)
where
e = electronic charge = [tex]1.6\times 10^{- 19} C[/tex]
From eqn (1) and (2):
[tex]p_{p} = \sqrt{2mK.E}[/tex] (3)
From eqn (2) and (3):
[tex]p_{p} = \sqrt{2meV}[/tex]
From the de-Broglie relation:
[tex]\lambda_{p} = \frac{h}{p_{p}}[/tex] (4)
where
[tex]\lambda_{p}[/tex] = wavelength of particle
h = Planck's constant
From eqn (3) and (4):
[tex]\lambda_{p} = \frac{h}{\sqrt{2meV}}[/tex]
