In a collision that is not perfectly elastic, what happens to the mechanical energy of the system?
1 answer:
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Explanation:
Here is the answer,
Lets be BF
Answer:
The acceleration of the electron is 1.457 x 10¹⁵ m/s².
Explanation:
Given;
initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s
distance traveled by the electron, d = 0.01 m
final velocity of the electron, v = 5.4 x 10⁶ m/s
The acceleration of the electron is calculated as;
v² = u² + 2ad
(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a
(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²
(2 x 0.01)a = 2.91375 x 10¹³
Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².
Answer:
Explanation:
The relation between the de Broglie wavelength and the momentum of the particle is given by
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
.... (1)
The de broglie wavelength of electron is given by
.... (2)
Divide equation (2) by equation (1), we get
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.