Advertisements
Advertisements
प्रश्न
Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity (ΔV) is equal to 1/4π of its velocity (V).
उत्तर
`∆"x".∆"p" ≥ "h"/(4π)`
`λ.("m"∆"v") ≥ "h"/(4π)`
∆v `≥ "h"/(4π("m"λ))`
∆v `≥ "h"/(4π xx "m" xx "h"/"mv")` .........[∵ λ = "h"/"mv"]
∆v `≥ "v"/(4π)`
Therefore, minimum uncertainty in velocity = `"v"/(4π)`
APPEARS IN
संबंधित प्रश्न
For d-electron, the orbital angular momentum is
If uncertainty in position and momentum are equal, then minimum uncertainty in velocity is
A macroscopic particle of mass 100 g and moving at a velocity of 100 cm s–1 will have a de Broglie wavelength of
The ratio of de Broglie wavelengths of a deuterium atom to that of an α - particle, when the velocity of the former is five times greater than that of later, is
How fast must a 54 g tennis ball travel in order to have a de Broglie wavelength that is equal to that of a photon of green light 5400 Å?
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the nucleus.
What is the de Broglie wavelength (in cm) of a 160 g cricket ball travelling at 140 Km hr−1?
Suppose that the uncertainty in determining the position of an electron in an orbit is 0.6 Å. What is the uncertainty in its momentum?
What is the de Broglie wavelength of an electron, which is accelerated from the rest, through a potential difference of 100V?
