Lets use Dimensional analysis.

<h3>LHS:-</h3>
![\\ \sf\longmapsto s=\left[M^0LT^0\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20s%3D%5Cleft%5BM%5E0LT%5E0%5Cright%5D)
<h3>RHS</h3>

![\\ \sf\longmapsto \left[M^0LT^{-1}\right]\left[M^0L^0T^1\right]+\dfrac{1}{2}\left[M^0LT^{-2}\right]\left[M^0L^0T^1\right]^2](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5BM%5E0LT%5E%7B-1%7D%5Cright%5D%5Cleft%5BM%5E0L%5E0T%5E1%5Cright%5D%2B%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5BM%5E0LT%5E%7B-2%7D%5Cright%5D%5Cleft%5BM%5E0L%5E0T%5E1%5Cright%5D%5E2)
![\\ \sf\longmapsto L+\dfrac{1}{2}\left[LT^{-2}\right]\left[T^2\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20L%2B%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5BLT%5E%7B-2%7D%5Cright%5D%5Cleft%5BT%5E2%5Cright%5D)


![\\ \sf\longmapsto [L^1]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5BL%5E1%5D)
![\\ \sf\longmapsto \left[M^0LT^0\right]](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20%5Cleft%5BM%5E0LT%5E0%5Cright%5D)
LHS=RHS
Hence verified
Answer:
B
A
Explanation:
please mark as brainliest :)
Explanation:
The De-Broglie wavelength in terms of potential difference is given by:

Where,
h is Planck's constant
m is mass of charged particle
V is potential difference
e is the amount of charge
It means that the De-Broglie wavelength is inversely proportional to the mass.
Since, the mass of the proton is more than the mass of the electron. So, the De- Broglie wavelength of the electron is larger than proton.
We're happy that you're asking for the "displacement", because displacement is simply the straight-line distance between the start-point and end-point, and we don't care about any of the motions or gyrations along the way.
From the graph:
-- The location of the object at time-zero, when time begins, is 10 meters.
-- The location of the object after 6.0 seconds is 4 meters.
-- The distance between the start-point and end-point is
(final location) - (initial location)
-- So Displacement = (4 meters) - (10 meters)
<em>Displacement = -6 meters</em>